JAVA经典算法50题

【程序1】   题目:古典问题:有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子长到第三个月后每个月又生一对兔子,假如兔子都不死,问每个月的兔子总数为多少?

1.程序分析:兔子的规律为数列1,1,2,3,5,8,13,21....
具体分析如下:
f(1) = 1(第1个月有一对兔子)
f(2) = 1(第2个月还是一对兔子)
f(3) = 2(原来有一对兔子,第3个开始,每个月生一对兔子)
f(4) = 3(原来有两对兔子,有一对可以生育)
f(5) = 5(原来有3对兔子,第3个月出生的那对兔子也可以生育了,那么现在有两对兔子可以生育)
f(6) = 8(原来有5对兔子,第4个月出生的那对兔子也可以生育了,那么现在有3对兔子可以生育)
..............

由以上可以看出,第n个月兔子的对数为
f(n) = f(n - 1) + f(n - 2);
f(n-1)是上个月的兔子数量,是原来有的。
f(n-2)是可以生育的兔子数,即多出来的数量。第n-2个月开始后的第3个月是第n个月,此时第n-2个月时的兔子都可以生育了。

public class Demo01 {
    public static void main(String args[]) {
        for (int i = 1; i <= 20; i++) {
            System.out.println(f(i));
        }
    }

    public static int f(int x) {
         if (x == 1 || x == 2) {
             return 1;
        } else {
            return f(x - 1) + f(x - 2);
        }
    }
}

public class Demo01 {
    public static void main(String args[]) {
        math mymath = new math();
        for (int i = 1; i <= 20; i++) {
            System.out.println(mymath.f(i));
        }
    }
}

class math {
    public int f(int x) {
        if (x == 1 || x == 2) {
            return 1;
        } else {
            return f(x - 1) + f(x - 2);
        }
    }
}


【程序2】   题目:判断101-200之间有多少个素数,并输出所有素数。

1.程序分析:判断素数的方法:用一个数分别去除2到sqrt(这个数),如果能被整除,则表明此数不是素数,反之是素数。

public class Demo02 {
    public static void main(String[] args) {
        for(int i = 2;i <= 200;i++) {
            boolean flag = true;
            for(int j = 2;j < i; j++) {
                if(i % j == 0) {
                    flag=false;
                    break;
                }
            }
            if(flag == true) {
                System.out.print(" "+i);
            }
        }
    }
}


【程序3】   题目:打印出所有的 水仙花数 ,所谓 水仙花数 是指一个三位数,其各位数字立方和等于该数本身。例如:153是一个 水仙花数 ,因为153=1的三次方+5的三次方+3的三次方。

1.程序分析:利用for循环控制100-999个数,每个数分解出个位,十位,百位。
public class Demo03 {
    public static void main(String args[]) {
        math mymath = new math();
        for (int i = 100; i <= 999; i++) {
            if (mymath.shuixianhua(i) == true) {
                System.out.println(i);
            }
        }
    }
}

class math {
    public boolean shuixianhua(int x) {
        int i = 0, j = 0, k = 0;
        i = x / 100;
        j = (x % 100) / 10;
        k = x % 10;
        if (x == i * i * i + j * j * j + k * k * k) {
            return true;
        } else {
           return false;
        }
    }
}


【程序4】   题目:将一个正整数分解质因数。例如:输入90,打印出90=2*3*3*5。 

1.程序分析:对n进行分解质因数,应先找到一个最小的质数i,然后按下述步骤完成:
(1)如果这个质数恰等于n,则说明分解质因数的过程已经结束,打印出即可。
(2)如果n > i,但n能被i整除,则应打印出i的值,并用n除以i的商,作为新的正整数你,重复执行第一步。
(3)如果n不能被i整除,则用i+1作为i的值,重复执行第一步。

import java.util.Scanner;
public class Demo04 {
    public Demo04() {
        super();
    }

    public void fenjie(int n) {
        for (int i = 2; i <= n; i++) {
            if (n % i == 0) {
                System.out.print(i);
                if (n != i) {
                    System.out.print("*");
                }
                fenjie(n/i);
            }
        }
        System.exit(0); // 退出程序
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.println("请输入N的值:");
        int N = in.nextInt();
        System.out.print( "分解质因数:" + N +"=");
        new Demo04().fenjie(N);
    }
}


【程序5】   题目:利用条件运算符的嵌套来完成此题:学习成绩=90分的同学用A表示,60-89分之间的用B表示,60分以下的用C表示。

1.程序分析:(a>b)?a:b这是条件运算符的基本例子。

import java.util.Scanner;
public class Demo05 {
    public static void main(String[] args) {
        System.out.println("请输入N的值:");
        Scanner in = new Scanner(System.in);
        int N = in.nextInt();
        System.out.println(N >= 90 ?"A": (N >= 60 ? "B":"C"));
    }
}


【程序6】   题目:输入两个正整数m和n,求其最大公约数和最小公倍数。

1.程序分析:利用辗除法。

import java.util.Scanner;
public class Demo06 {
    public static void main(String[] args) {
        int a,b,m,n;
        Scanner in = new Scanner(System.in);
        System.out.println("请输入一个正整数:");
        a = in.nextInt();
        System.out.println("再输入一个正整数:");
        b = in.nextInt();
        commonDivisor use = new commonDivisor();
        m = use.commonDivisor(a,b);
        n = a*b/m;
        System.out.println("最大公约数:"+m);
        System.out.println("最小公倍数:"+n);
    }
}

class commonDivisor {
    public int commonDivisor(int x,int y) {
        if(x < y) {
            int t = x;
            x = y;
            y = t;  
        }
        
        while(y != 0) {
            if(x == y) {
                return x;
            } else {
                int k = x % y;
                x = y;
                y = k;
            }
        }
        return x;
    }
}


【程序7】   题目:输入一行字符,分别统计出其中英文字母、空格、数字和其它字符的个数。

1.程序分析:利用for循环语句,if条件语句。

import java.util.Scanner;
public class Demo07 {
    public static void main(String[] args) {
        System.out.println("请输入一个字符串;");
        Scanner in = new Scanner(System.in);
        String str = in.nextLine();
        char[] ch = str.toCharArray();
        count use = new count();
        use.count(ch);
    }
}

class count{
    int digital,character,blank,other;
    public void count(char[] arr) {
        for(int i = 0; i < arr.length(); i++) {
            if (arr[i] >= '0' && arr[i] <= '9') {
                digital++;
            } else if ((arr[i] >= 'a' && arr[i] <= 'z') || (arr[i] >= 'A' && arr[i] <= 'Z')) {
                character++;
            } else if (arr[i] == ' ') {
                blank++;
            } else {
                other++;
            }
        }
        System.out.println("数字个数:"+digital);
        System.out.println("英文字母个数:"+character);
        System.out.println("空格个数:"+blank);
        System.out.println("其他字符个数:"+other);
    }
}


【程序8】   题目:求s = a + aa + aaa + aaaa + aa...a的值,其中a是一个数字。例如2 + 22 + 222 + 2222 + 22222(此时共有5个数相加),几个数相加有键盘控制。

1.程序分析:关键是计算出每一项的值。

import java.util.Scanner;
public class Demo08 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.println(请输入a的值);
        int a = in.nextInt();
        System.out.println(请输入n个数);
        int n = in.nextInt();
        int s = 0,t=0;
        for (int i = 1; i <= n; i++) {
            t += a;
            a = a*10;
            s += t;
        }
        System.out.println(s);
    }
}


【程序9】   题目:一个数如果恰好等于它的因子之和,这个数就称为"完数"。例如6=1+2+3。编程找出1000以内的所有完数。

public class Demo09 {
    public static void main(String[] args) {
        int s;
        for (int i = 1; i <= 1000; i++) {
            s = 0;
            for (int j = 1; j < i; j++) {
                if (i % j == 0) {
                    s = s + j;
                }
            }
            if (s == i)  {
                System.out.print(i + " " );
            }
        }
        System.out.println();
    }
}


【程序10】 题目:一球从100米高度自由落下,每次落地后反跳回原高度的一半;再落下,求它在第10次落地时,共经过多少米?第10次反弹多高?

public class Demo10 {
    public static void main(String[] args) {
        double s = 0;
        double h = 100;
        for (int i = 1; i <= 10; i++) {
            s += h;
            h = h/2;
            s += h;
        }
        System.out.println("经过路程:"+s);
        System.out.println("反弹高度:"+h);
    }
}


【程序11】   题目:有1、2、3、4个数字,能组成多少个互不相同且无重复数字的三位数?都是多少?

1.程序分析:可填在百位、十位、个位的数字都是1、2、3、4。组成所有的排列后再去掉不满足条件的排列。   

public class Demo11 {
    public static void main(String[] args) {
        int count = 0;
        for (int i = 1; i <= 4; i++) {
            for (int j = 1; j <= 4; j++) {
                for (int k = 1; k <= 4; k++) {
                    if (i != j && j != k && i != k) {
                        count += 1;
                        System.out.println(i * 100 + j * 10 + k);
                    }
                }
            }
        }
        System.out.println("共" + count + "个三位数");
    }
}


【程序12】  题目:企业发放的奖金根据利润提成。利润(I)低于或等于10万元时,奖金可提10%;利润高于10万元,低于20万元时,低于10万元的部分按10%提成,高于10万元的部分,可提成7.5%;20万到40万之间时,高于20万元的部分,可提成5%;40万到60万之间时高于40万元的部分,可提成3%;60万到100万之间时,高于60万元的部分,可提成1.5%,高于100万元时,超过100万元的部分按1%提成,从键盘输入当月利润lirun,求应发放奖金总数sum?

1.程序分析:请利用数轴来分界,定位。注意定义时需把奖金定义成长整型。

import java.util.Scanner;
public class Demo12 {
    public static void main(String[] args) {
        double sum;
        System.out.println("输入当月利润:(万元)");
        Scanner in = new Scanner(System.in);
        double lirun = in.nextDouble();
        if (lirun <= 10) {
            sum = lirun * 0.1;
        } else if (lirun <= 20) {
            sum = 10*0.1 + (lirun - 10) * 0.075;
        } else if (lirun <= 40) {
            sum = 10*0.1 + 10*0.075 + (lirun - 20) * 0.05;
        } else if (lirun <= 60) {
            sum = 10*0.1 + 10*0.075 + 10*0.05 + (lirun - 40) * 0.03;
        } else if (lirun <= 100) {
            sum = 10*0.1 + 10*0.075 + 10*0.05 + 10*0.03 + (lirun - 60) * 0.015;
        } else {
            sum = 10*0.1 + 10*0.075 + 10*0.05 + 10*0.03 + 10*0.015 + (lirun - 100) * 0.01;
        }
        System.out.println("应发的奖金是:"+sum+"(万元)");
    }
}


【程序13】   题目:一个整数,它加上100后是一个完全平方数,加上168又是一个完全平方数,请问该数是多少?

1.程序分析:在10万以内判断,先将该数加上100后再开方,再将该数加上168后再开方,如果开方后的结果满足如下条件,即是结果。请看具体分析:

public class Demo13 {
    public static void main(String[] args) {
        for(int x = 1 ; x < 100000 ; x ++) {
            int value1 = Math.sqrt(x + 100);
            int value2 = Math.sqrt(x + 100 + 168)
            if (value1 * value1 == x && value2 * value2 == x) {
                System.out.println(x+"加上100后是一个完全平方数,加上168又是一个完全平方数");
            }
        }
    }
}


【程序14】 题目:输入某年某月某日,判断这一天是这一年的第几天?

1.程序分析:以3月5日为例,应该先把前两个月的加起来,然后再加上5天即本月的第几天,特殊情况,闰年且输入月份大于3时需考虑多加一天。

import java.util.Calendar;
import java.util.Scanner;
public class Demo14 {
    public static void main(String[] args) {
        System.out.println("请输入年,月,日:");
        Scanner in = new Scanner(System.in);
        int year = in.nextInt();
        int month = in.nextInt();
        int day = in.nextInt();
        Calendar cal = Calendar.getInstance();
        cal.set(year, month - 1, day);
        int sum = cal.get(Calendar.DAY_OF_YEAR);
        System.out.println("这一天是这一年的第" + sum +"天");
    }
}

import java.util.*;
public class Demo14 {
    public static void main(String[] args) {
        int year,month,day,sum = 0;
        Scanner in = new Scanner(System.in);
        System.out.println("输入年:");
        year = in.nextInt();
        System.out.println("输入月:");
        month = in.nextInt();
        System.out.println("输入日:");
        day = in.nextInt();
        switch(month) {
            case 1:
                sum=0;
                break;
            case 2:
                sum=31;
                break;
            case 3:
                sum=59;
                break;
            case 4:
                sum=90;
                break;
            case 5:
                sum=120;
                break;
            case 6:
                sum=151;
                break;
            case 7:
                sum=181;
                break;
            case 8:
                sum=212;
                break;
            case 9:
                sum=243;
                break;
            case 10:
                sum=273;
                break;
            case 11:
                sum=304;
                break;
            case 12:
                sum=334;
                break;
            default:
                System.out.println("wrong input!");
                return;
        }
        sum = sum+day;
        boolean leap;
        if(year % 400 == 0 || (year % 4 == 0 && year % 100 != 0)) {
            leap = true;
        } else {
            leap = false;
        }
        if(leap && month > 2) {
            sum++;
        }
        System.out.println("It is the "+sum+"th day.");
    }
}

import java.util.Scanner;
public class Demo14 {
    public static void main(String[] args) {
        System.out.println("请输入年 月 日:");
        Scanner in = new Scanner(System.in);
        int year = in.nextInt();
        int month = in.nextInt();
        int day = in.nextInt();
        System.out.println("是该年的第"+count(year,month,day)+"天");
    }

    public static int count(int year,int month,int day) {
        int sum = 0;
        int days = 0;
        for(int i = 1;i < 3; i++) {
             switch(i) {
                case 1:
                case 3:
                case 5:
                case 7:
                case 8:
                case 10:
                case 12:
                    days=31;
                    break;
                case 4:
                case 6:
                case 9:
                case 11:
                    days=30;
                    break;
                case 2:
                    if(year % 400 == 0 || year % 4 == 0 && year % 100 != 0) {
                        days=29;
                    } else {
                        days=28;
                    }
                    break;
             }
            sum+=days;
        }
        sum+=day;
        return sum;
     }
}


【程序15】 题目:输入三个整数x,y,z,请把这三个数由小到大输出。

1.程序分析:我们想办法把最小的数放到x上,先将x与y进行比较,如果x>y则将x与y的值进行交换,然后再用x与z进行比较,如果x>z则将x与z的值进行交换,这样能使x最小。

import java.util.Arrays;
import java.util.Scanner;
public class Demo15 {
    public static void main(String[] args) {
        System.out.print("请输入三个数:");
        Scanner in = new Scanner(System.in);
        int[] arr = new int[3];
        for (int i = 0; i < 3; i++) {
            arr[i] = in.nextInt();
        }
        Arrays.sort(arr);
        for (int i = 0 ; i < arr.length() ; i++) {
            System.out.print(arr[i] + " ");
        }
    }
}

if(x > y) {
    int t = x;
    x = y;
    y = t;
}
if(x > z) {
    int t = x;
    x = z;
    z = t;
}
if(y > z) {
    int t = y;
    y = z;
    z = t;
}


【程序16】 题目:输出9*9口诀乘法表。

1.程序分析:分行与列考虑,共9行9列,i控制行,j控制列。

出现重复的乘积(全矩形)
public class Demo16 {
    public static void main(String[] args) {
        for (int i = 1; i <= 9; i++) {
            for (int j = 1; j <= 9; j++) {
                System.out.print(i + "*" + j + "=" + (i*j) + "\t");
            }
            System.out.println();
        }
    }
}

不现重复的乘积(下三角)
public class Demo16 {
    public static void main(String[] args) {
        for (int i = 1; i <= 9; i++) {
            for (int j = 1; j <= i; j++) {
                System.out.print(i + "*" + j + "=" + (i*j) + "\t");
            }
            System.out.println();
        }
    }
}


【程序17】   题目:猴子吃桃问题:猴子第一天摘下若干个桃子,当即吃了一半,还不瘾,又多吃了一个第二天早上又将剩下的桃子吃掉一半,又多吃了一个。以后每天早上都吃了前一天剩下的一半零一个。到第10天早上想再吃时,见只剩下一个桃子了。求第一天共摘了多少。

1.程序分析:采取逆向思维的方法,从后往前推断。

public class Demo17 {
    public static void main(String[] args) {
        int sum = 1;
        for (int i = 0; i < 9; i++) {
            sum = (sum + 1) * 2;
        }
        System.out.println("第一天共摘"+sum);
    }
}


【程序18】   题目:两个乒乓球队进行比赛,各出三人。甲队为a,b,c三人,乙队为x,y,z三人。已抽签决定比赛名单。有人向队员打听比赛的名单。a说他不和x比,c说他不和x,z比,请编程序找出三队赛手的名单。

public class Demo18 {
    static char[] m = { 'a', 'b', 'c' };
    static char[] n = { 'x', 'y', 'z' };

    public static void main(String[] args) {
        for (int i = 0; i < m.length; i++) {
            for (int j = 0; j < n.length; j++) {
                if (m[i] == 'a' && n[j] == 'x') {
                    continue;
                } else if (m[i] == 'a' && n[j] == 'y') {
                    continue;
                } else if ((m[i] == 'c' && n[j] == 'x') || (m[i] == 'c' && n[j] == 'z')) {
                    continue;
                } else if ((m[i] == 'b' && n[j] == 'z') || (m[i] == 'b' && n[j] == 'y')) {
                    continue;
                } else {
                    System.out.println(m[i] + " vs " + n[j]);
                }
            }
        }
    }
}

public class Demo18 {
    public String a, b, c;

    public Demo18(String a, String b, String c) {
        this.a = a;
        this.b = b;
        this.c = c;
    }

    public static void main(String[] args) {
        Demo18 arr_a = new Demo18("a", "b", "c");
        String[] b = { "x", "y", "z" };
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                for (int k = 0; k < 3; k++) {
                    Demo18 arr_b = new Demo18(b[i], b[j], b[k]);
                    if (!arr_b.a.equals(arr_b.b) & !arr_b.b.equals(arr_b.c)
                        & !arr_b.c.equals(arr_b.a) & !arr_b.a.equals("x")
                        & !arr_b.c.equals("x") & !arr_b.c.equals("z")) {
                        System.out.println(arr_a.a + "--" + arr_b.a);
                        System.out.println(arr_a.b + "--" + arr_b.b);
                        System.out.println(arr_a.c + "--" + arr_b.c);
                    }
                }
            }
        }
    }
}


【程序19】  题目:打印出如下图案(菱形)

1.程序分析:先把图形分成两部分来看待,前四行一个规律,后三行一个规律,利用双重for循环,第一层控制行,第二层控制列。

三角形:
*
***
******
********
******
***
*

public class Demo19 {
    public static void main(String[] args) {
        int i = 0;
        int j = 0;
        for ( i = 1; i <= 4; i++) {
            for ( j = 1; j <= 2 * i - 1; j++) {
                System.out.print("*");
            }
            System.out.println();
        }

        for ( i = 3; i >= 1; i--) {
            for ( j = 1; j <= 2 * i - 1; j++) {
                System.out.print("*");
            }
            System.out.println();
        }
    }
}

菱形:
     *
    ***
   *****
  *******
   *****
    ***
     *

public class Demo19 {
    public static void main(String[] args) {
        int i = 0;
        int j = 0;
        for (i = 1; i <= 4; i++)  {
            for (int k = 1; k <= 4 - i; k++) {
                System.out.print( " " );
            }

            for (j = 1; j <= 2 * i - 1; j++) {
                System.out.print("*");
            }
            System.out.println();
        }

        for (i = 3; i >= 1; i--) {
            for (int k = 1; k <= 4 - i; k++) {
                System.out.print( " " );
            }
        for (j = 1; j <= 2 * i - 1; j++) {
            System.out.print("*");
        }
        System.out.println();
        }
    }
}


【程序20】   题目:有一分数序列:2/1,3/2,5/3,8/5,13/8,21/13...求出这个数列的前20项之和。

1.程序分析:请抓住分子与分母的变化规律。   

public class Demo20 {
    public static void main(String[] args) {
        float fm = 1.0f;
        float fz = 1.0f;
        float temp;
        float sum = 0f;
        for (int i = 0; i < 20; i++) {
            temp = fm;
            fm = fz;
            fz = fz + temp;
            System.out.println((int) fz + "/" + (int) fm);
            sum += fz / fm;
        }
        System.out.println(sum);
    }
}


【程序21】   题目:求1+2!+3!+...+20!的和。

1.程序分析:此程序只是把累加变成了累乘。

public class Demo21 {
    public static void main(String[] args) {
        long sum = 0;
        long fac = 1;
        for (int i = 1; i <= 20; i++) {
            fac = fac * i;
            sum += fac;
        }
        System.out.println(sum);
    }
}


【程序22】   题目:利用递归方法求5!。

1.程序分析:递归公式:f(n)=f(n-1)*4!

import java.util.Scanner;
public class Demo22 {
    public static long fac(int n) {
        long value = 0;
        if (n == 1 || n == 0) {
            value = 1;
        } else if (n > 1) {
            value = n * fac(n - 1);
        }
        return value;
    }

    public static void main(String[] args) {
        System.out.println("请输入一个数:");
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        System.out.println(n + "的阶乘为:" + fac(n));
    }
}


【程序23】   题目:有5个人坐在一起,问第五个人多少岁?他说比第4个人大2岁。问第4个人岁数,他说比第3个人大2岁。问第三个人,又说比第2人大两岁。问第2个人,说比第一个人大两

岁。最后问第一个人,他说是10岁。请问第五个人多大?

1.程序分析:利用递归的方法,递归分为回推和递推两个阶段。要想知道第五个人岁数,需知道第四人的岁数,依次类推,推到第一人(10岁),再往回推。

直接求解:

public class Demo23 {
    public static void main(String[] args) {
        int n = 10;
        for (int i = 0; i < 4; i++) {
            n = n + 2;
        }
        System.out.println("第五个人" + n + "岁");
    }
}

递归求解:

public class Demo23 {
    public static int getAge(int n) {
        if (n == 1) {
            return 10;
        }
        return 2 + getAge(n - 1);
    }

    public static void main(String[] args) {
        System.out.println("第五个的年龄为" + getAge(5));
    }
}


【程序24】   题目:给一个不多于5位的正整数,要求:一、求它是几位数,二、逆序打印出各位数字。 

本题原方法:

import java.util.Scanner;
public class Demo24 {
    public static void main(String[] args) {
        Demo24 use = new Demo24();
        System.out.println("请输入:");
        Scanner in = new Scanner(System.in);
        long a = in.nextLong();
        if (a < 0 || a >= 100000) {
            System.out.println("Error Input, please run this program Again!");
            System.exit(0);
        }

        if (a >= 0 && a <= 9) {
            System.out.println(a + "是一位数");
            System.out.println("按逆序输出是:"  + a);
        } else if (a >= 10 && a <= 99) {
            System.out.println(a + "是二位数");
            System.out.println("按逆序输出是:");
            use.converse(a);
        } else if (a >= 100 && a <= 999) {
            System.out.println(a + "是三位数");
            System.out.println("按逆序输出是:");
            use.converse(a);
        } else if (a >= 1000 && a <= 9999) {
            System.out.println(a + "是四位数");
            System.out.println("按逆序输出是:");
            use.converse(a);
        } else if (a >= 10000 && a <= 99999) {
            System.out.println(a + "是五位数");
            System.out.println("按逆序输出是:");
            use.converse(a);
        }
    }

    public void converse(long l) {
        String s = Long.toString(l);
        char[] ch = s.toCharArray();
        for (int i = ch.length - 1; i >= 0; i--) {
            System.out.print(ch[i]);
        }
    }
}

个人版方法:

import java.util.Scanner;
public class Demo24 {
    public static void main(String[] args) {
        System.out.println("请输入:");
        Scanner in = new Scanner(System.in);
        String str = in.next();
        if (str.matches("\\d+")) { //正则表达式
            System.out.println("输入的是" + str.length() + "位数");
            StringBuffer buf = new StringBuffer(str);
            System.out.println(buf.reverse());//字符串反转
        }
    }
}


【程序25】   题目:一个5位数,判断它是不是回文数。即12321是回文数,个位与万位相同,十位与千位相同。

原方法:

import java.util.Scanner;
public class Demo25 {
    static int[] a = new int[5];
    static int[] b = new int[5];
    public static void main(String[] args) {
        boolean is = false;
        System.out.println("Please input:");
        Scanner in = new Scanner(System.in);
        long l = in.nextLong();
        if (l > 99999 || l < 10000) {
            System.out.println("Input error, please input again!");
            l = in.nextLong();
        }

        for (int i = 4; i >= 0; i--) {
            a[i] = (int) (l / (long) Math.pow(10, i));
            l = (l % (long) Math.pow(10, i));
        }
        System.out.println();
        for (int i = 0, j = 0; i < 5; i++, j++) {
            b[j] = a[i];
        }
        for (int i = 0, j = 4; i < 5; i++, j--) {
            if (a[i] != b[j]) {
                is = false;
                break;
            } else {
                is = true;
            }
        }

        if (is == false) {
            System.out.println("is not a Palindrom!");
        } else if (is == true) {
            System.out.println("is a Palindrom!");
        }
    }
}

个人版:

import java.util.Scanner;
public class Demo25 {
    public static void main(String[] args) {
        System.out.println("请输入:");
        Scanner in = new Scanner(System.in);
        String str = in.next();
        int l = Integer.parseInt(str);//转换成整数
        if (l < 10000 || l > 99999) {
            System.out.println("输入错误!");
            System.exit(0);
        }
        boolean is=false;
        char[] ch = str.toCharArray();

        for(int i=0; i < ch.length/2; i++) {
            if(ch[i] != ch[ch.length-i-1]){
                is = false;
            } else {
                is = true;
            }
        }

        if(is) {
            System.out.println("这是一个回文!");
        } else {
            System.out.println("不是一个回文!");
        }
     }
}


【程序26】   题目:请输入星期几的第一个字母来判断一下是星期几,如果第一个字母一样,则继续判断第二个字母。

1.程序分析:用情况语句比较好,如果第一个字母一样,则判断用情况语句或if语句判断第二个字母。

import java.util.Scanner;
public class Demo26 {
    public static void main(String[] args) {
        char weekSecond;//保存第二字母
        Scanner in = new Scanner(System.in);//接收用户输入
        System.out.println("请输入星期的第一个字母:");
        String letter = in.next();
        if (letter.length() == 1) {//判断用户控制台输入字符串长度是否是一个字母
            char weekFirst = letter.charAt(0);//取第一个字符
            switch (weekFirst) {
                case 'm':
                case 'M':
                    System.out.println("星期一(Monday)");
                    break;
                case 't':
                case 'T':
                System.out.print("由于星期二(Tuesday)与星期四(Thursday)均以字母T开头,故需输入第二个字母才能正确判断:");
                letter = in.next();
                if (letter.length() == 1) {
                    weekSecond = letter.charAt(0);
                    if (weekSecond == 'U' || weekSecond == 'u') {
                        System.out.println("星期二(Tuesday)");
                        break;
                    } else if (weekSecond == 'H' || weekSecond == 'h') {
                        System.out.println("星期四(Thursday)");
                        break;
                    } else {
                        System.out.println("Error!");
                        break;
                    }
                } else {
                    System.out.println("输入错误,只能输入一个字母,程序结束!");
                    break;
                }

           case 'w':
           case 'W':
                System.out.println("星期三(Wednesday)");
                break;
            case 'f':
            case 'F':
                System.out.println("星期五(Friday)");
                break;
            case 's':
            case 'S':
                System.out.print("由于星期六(Saturday)与星期日(Sunday)均以字母S开头,故需输入第二个字母才能正确判断:");
                letter = in.next();
                if (letter.length() == 1) {
                    weekSecond = letter.charAt(0);
                    if (weekSecond == 'A' || weekSecond == 'a') {
                        System.out.println("星期六(Saturday)");
                        break;
                    } else if (weekSecond == 'U' || weekSecond == 'u') {
                        System.out.println("星期日(Sunday)");
                        break;
                    } else {
                        System.out.println("Error!");
                        break;
                    }
                } else {
                    System.out.println("输入错误,只能输入一个字母,程序结束!");
                    break;
                }

            default:
                System.out.println("输入错误,不能识别的星期值第一个字母,程序结束!");
                break;
           }
        } else {
            System.out.println("输入错误,只能输入一个字母,程序结束!");
        }
    }
}


【程序27】   题目:求100之内的素数

public class Demo27{
    public static void main(String args[]) {
        int w = 1;
        for(int i = 2; i <= 100; i++) {
            for(int j = 2; j < i; j++) {
                f(i % j == 0) {
                    break;
                }
            }
            System.out.println(i+"是素数");
        }
    }
}


【程序28】   题目:对10个数进行排序。

1.程序分析:可以利用选择法,即从后9个比较过程中,选择一个最小的与第一个元素交换,下次类推,即用第二个元素与后8个进行比较,并进行交换。

本例代码为生成随机10个数排序,并输入1个数,插入重排序输出:

import java.util.Arrays;
import java.util.Random;
import java.util.Scanner;
public class Demo28 {

    public static void main(String[] args) {
        int arr[] = new int[11];
        Random r = new Random();
        for (int i = 0; i < 10; i++) {
            arr[i] = r.nextInt(100) + 1; //得到10个100以内的整数
        }
        Arrays.sort(arr);
        for (int i = 0; i < arr.length; i++) {
            System.out.print(arr[i] +"\t");
        }
        System.out.print("\nPlease Input a int number:" );

        Scanner in = new Scanner(System.in);
        arr[10] = in.nextInt();
        Arrays.sort(arr);
        for (int i = 0; i < arr.length; i++) {
            System.out.print(arr[i] +"\t");
        }
    }
}

个人代码:

import java.util.Arrays;
import java.util.Scanner;
public class Demo28 {
    public static void main(String[] args) {
        System.out.println("请输入10个数:");
        Scanner in = new Scanner(System.in);
        int[] arr = new int[10];
        for (int i = 0; i < 10; i++) {
            arr[i] = in.nextInt();
        }
        System.out.println("原数组为:");
        for (int x : arr) {  //foreach遍历
            System.out.print( x + "\t");
        }
        Arrays.sort(arr);
        System.out.println();
        System.out.println("排序后为:");
        for (int i = 0; i < arr.length; i++) {
            System.out.print(arr[i] + "\t");
        }
    }
}


【程序29】   题目:求一个3*3矩阵主对角线元素之和。

1.程序分析:利用双重for循环控制输入二维数组,再将a[i][i]累加后输出。

public class Demo29 {
    public static void main(String[] args) {
        double sum = 0;
        int array[][] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 7, 8 } };
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                if (i == j) {
                    sum = sum + array[i][j];
                }
            }
            System.out.println(sum);
        }
    }
}


【程序30】   题目:有一个已经排好序的数组。现输入一个数,要求按原来的规律将它插入数组中。

1.程序分析:首先判断此数是否大于最后一个数,然后再考虑插入中间的数的情况,插入后此元素之后的数,依次后移一个位置。

import java.util.Random;
public class Demo30 {
    public static void main(String[] args) {
        int temp = 0;
        int arr[] = new int[12];
        Random r = new Random();
        for (int i = 0; i <= 10; i++) {
            arr[i] = r.nextInt(1000);
        }
        for (int i = 0; i <= 10; i++) {
            System.out.print(arr[i] + "\t");
        }
        for (int i = 0; i <= 9; i++)  {
            for (int k = i + 1; k <= 10; k++) {
                if (arr[i] > arr[k]) {
                    temp = arr[i];
                    arr[i] = arr[k];
                    arr[k] = temp;
                }
            }
        }
        System.out.println();
        for (int k = 0; k <= 10; k++) {
            System.out.print(arr[k] + "\t");
        }
        arr[11] = r.nextInt(1000);
        for (int k = 0; k <= 10; k++) {
            if (arr[k] > arr[11]) {
                temp = arr[11];
                for (int j = 11; j >= k + 1; j--) {
                    arr[j] = arr[j - 1];
                    arr[k] = temp;
                }
            }
        }
        System.out.println();
        for (int k = 0; k <= 11; k++) {
            System.out.print(arr[k] + "\t");
        }
    }
}


【程序31】   题目:将一个数组逆序输出。

用逆序循环控制变量输出:

public class Demo31 {
    public static void main(String[] args) {
        int[] a = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 };
        for (int i = a.length - 1; i >= 0; i--) {
            System.out.print(a[i] + " ");
        }
    }
}


【程序32】   题目:取一个整数a从右端开始的第4~7位数字。

import java.util.*;
public class Demo32 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.print("请输入一个7位以上的正整数:");
        long l = in.nextLong();
        String str = Long.toString(l);
        char[] ch = str.toCharArray();
        int j=ch.length;
        if (j < 7) {
            System.out.println("输入错误!");
        } else {
            System.out.println("截取从右端开始的4~7位是:"+ch[j-7]+ch[j-6]+ch[j-5]+ch[j-4]);
        }
    }
}


【程序33】   题目:打印出杨辉三角形(要求打印出10行如下图)

1.程序分析:
1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10   10   5   1

public class Demo33 {
    public static void main(String args[]) {
        int i, j;
        int a[][];
        int n = 10;
        a = new int[n][n];
        for (i = 0; i < n; i++) {
            a[i][i] = 1;
            a[i][0] = 1;
        }
        for (i = 2; i < n; i++) {
            for (j = 1; j <= i - 1; j++) {
                a[i][j] = a[i - 1][j - 1] + a[i - 1][j];
            }
        }
        for (i = 0; i < n; i++) {
            for (j = 0; j <= i; j++) {
                System.out.printf(a[i][j] + "\t");
            }
            System.out.println();
        }
    }
}


【程序34】   题目:输入3个数a,b,c,按大小顺序输出。
import java.util.Arrays;
public class Demo34 {
    public static void main(String[] args) {
        int[] arrays = { 800, 56, 500 };
        Arrays.sort(arrays);
        for (int n = 0; n < arrays.length; n++)
            System.out.println(arrays[n]);
    }
}


【程序35】   题目:输入数组,最大的与第一个元素交换,最小的与最后一个元素交换,输出数组。

import java.util.*;
public class Demo35 {
    public static void main(String[] args) {
        int i, min = 0, max = 0, n, temp1, temp2;
        int a[];
        System.out.println("定义数组的长度:");
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        a = new int[n];
        for (i = 0; i < n; i++) {
            System.out.print("输入第" + (i + 1) + "个数据:");
            a[i] = in.nextInt();
        }
        for (i = 1; i < n; i++) {
            if (a[i] > a[max]) {
                max = i;
            }
            if (a[i] < a[min]) {
                min = i;
            }
        }
        temp1 = a[0];
        a[0] = a[max];
        a[max] = temp1;
        temp2 = a[min]; 
        if (min != 0) { // 如果最小值不是a[0],执行下面
            a[min] = a[n - 1];
            a[n - 1] = temp2; 
        } else {  //如果最小值是a[0],执行下面
            a[max] = a[n - 1];
            a[n - 1] = temp1;
        }
        for (i = 0; i < n; i++) { 
            System.out.print(a[i] + " " );
        }
    }
}


【程序36】   题目:有n个整数,使其前面各数顺序向后移m个位置,最后m个数变成最前面的m个数

import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;
public class Demo36 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.println("输入数字个数n:");
        int n = in.nextInt();
        System.out.println("输入后移位数m:");
        int m = in.nextInt();
        LinkedList list = new LinkedList();
        for (int i = 0; i < n; i++) {
            System.out.println("请输入第"+(i+1)+"个数:");
            list.add(in.nextInt());
        }
        System.out.println("原数据排序为:");
        for (int t : list) {
            System.out.print(t + " " );
        }
        System.out.println();
        List temp1 = list.subList(list.size() - m, list.size());        
        List temp2 = list.subList(0, list.size() - m);
        temp2.addAll(0, temp1);
        System.out.println("移动后排序为;");
        for (int t : temp2) {
            System.out.print(t + " " );
        }
    }
}

import java.util.*;
public class Demo36 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);       
        System.out.println("请定义数组的长度:");
        int n = in.nextInt();
        System.out.println("请输入移动的位数:");
        int m = in.nextInt();
        int [] arr = new int [n];
        int [] brr = new int [n];
        for(int i = 0;i < n;i ++) {
            System.out.println("请输入第"+(i+1)+"个数:");
            arr[i] = in.nextInt();
        }
        System.out.println("排序前:");
        for(int i = 0;i < n;i ++) {
            System.out.print(arr[i]+" ");
        }
        System.out.println();
        for(int i = 0;i < m;i ++) {
            brr[i] = arr[n - m + i];
        }
        for(int i = 0;i < n - m;i ++) {
            arr[m + i] = arr[i];
        }
        for(int i=0;i < m;i ++) {
            arr[i] = brr[i];
        }
        System.out.println("排序后:");
        for(int i = 0;i < m;i ++) {
            System.out.print(arr[i]+" ");
        }
    }
}


【程序37】   题目:有n个人围成一圈,顺序排号。从第一个人开始报数(从1到3报数),凡报到3的人退出圈子,问最后留下的是原来第几号的那位。
(约瑟夫环问题,百度百科有时间复杂度最简单的数学方法)
原例代码:

import java.util.Scanner;
public class Demo37 {
    public static void main(String[] args) {
        System.out.println("请输人数n:");
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        boolean[] arr = new boolean[n];
        for (int i = 0; i < arr.length; i++) {
            arr[i] = true; //下标为TRUE时说明还在圈里
        }
        int leftCount = n;
        int countNum = 0;
        int index = 0;
        while (leftCount > 1) {
            if (arr[index] == true) { //当在圈里时
                countNum++;  //报数递加
                if (countNum == 3) { //报数为3时
                    countNum = 0; //从零开始继续报数
                    arr[index] = false; //此人退出圈子
                    leftCount--; //剩余人数减一
                }
            }
            index++; //每报一次数,下标加一
            if (index == n) { //是循环数数,当下标大于n时,说明已经数了一圈,
                index = 0; //将下标设为零重新开始。
            }
        }
        for (int i = 0; i < n; i++) {
            if (arr[i] == true) {
                System.out.println(i);
            }
        }
    }
}

个人代码1:
import java.util.Scanner;
public class Demo37 {
    public static void main(String[] args) {
        System.out.println("请输入人数:");
        Scanner in = new Scanner(System.in);
        int[] a = new int[in.nextInt()];
        for (int i = 0; i < a.length; i++) {
            a[i] = 1;
        }
        int left = a.length;
        int j = 0;
        int num = 0;
        while (left > 1) {
            if (a[j] == 1) {
                num ++;
            }
`           if (num == 3) {
                a[j] = 0;
                num = 0;
                left--;
            }
            j++;
            if (j == a.length) {
                j = 0;
            }
        }
        for (int i = 0; i < a.length; i++) {
            if (a[i] == 1) {
                System.out.println("最后留下的人是"+ (i + 1) + "号");
                break;
            }
        }
    }
}

个人代码2:
import java.util.LinkedList;
import java.util.Scanner;
public class Demo37 {
    public static void main(String[] args) {
        LinkedList l = new LinkedList();
        System.out.println("请输入人数:");
        Scanner in = new Scanner(System.in);
        int len = in.nextInt();
        for (int i = 0; i < len; i++) {
            l.add(i + 1);
        }
        int sum = 0;
        int temp = 0;
        for (int i = 0; sum != len - 1;) {
            if (l.get(i) != 0) {
                temp++;
            }
            if (temp == 3) {
                l.remove(i);
                l.add(i, 0);
                temp = 0;
                sum++;
            }
            i++;
            if (i == l.size()) {
                i = 0;
            }
        }
        for (int t : l) {
            if (t != 0) {
                System.out.println("最后留下的人是" + t + "号");
            }
        }
    }
}


【程序38】   题目:写一个函数,求一个字符串的长度,在main函数中输入字符串,并输出其长度。

import java.util.Scanner;
public class Demo38 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.println("请输入一个字符串:");
        String mys = in.next();
        System.out.println(str_len(mys));
    }
    public static int str_len(String x) {
        return x.length();
    }
}


【程序39】  题目:编写一个函数,输入n为偶数时,调用函数求1/2+1/4+...+1/n,当输入n为奇数时,调用函数1/1+1/3+...+1/n

import java.util.Scanner;
public class Demo39 {
    public static double ouShu(int n) {
        double result = 0;
        for (int i = 2; i <= n; i = i + 2) {
            result +=  1 / (double) i;
        }
        return result;
    }
    public static double jiShu(int n) {
        double result = 0;
        for (int i = 1; i <= n; i = i + 2) {
            result += 1 / (double) i;
        }
        return result;
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.println("输入n的值:");
        int n = in.nextInt();
        if (n % 2 == 0) { //偶数,1/2+1/4+...+1/n
            System.out.println(ouShu(n));
        } else { //奇数,1/1+1/3+...+1/n
            System.out.println(jiShu(n));
        }
    }
}


【程序40】  题目:字符串排序。

(利用容器类中的sort方法)
import java.util.*;
public class Demo40 {
    public static void main(String[] args) {
        ArrayList list = new ArrayList(); 
        list.add("010102");
        list.add("010003");
        list.add("010201");
        Collections.sort(list);
        for (int i = 0; i < list.size(); i++) {
            System.out.println(list.get(i));
        }
    }
}

import java.util.*;
public class Demo40 {
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        System.out.println("请定义字符串的个数:");
        int n = n.nextInt();
        String[] str = new String[n];
        for(int i = 0; i < str.length; i++) {
            System.out.println("请输入第"+(i+1)+"字符串:");
            str[i] = in.next();
        }
        strSort(n,str);
        System.out.println("字符串排序后:");
        for(int i = 0; i < str.length; i++) {
            System.out.print(str[i]+" ");
        }
    }
    public static void strSort(int n,String[] arr) {
        for(int i=0; i < n; i++ ) {
            for(int j=i+1; j < n; j++) {
                if(compare(arr[i], arr[j]) == false) {
                    String temp = arr[i];
                    arr[i] = arr[j];
                    arr[j] = temp;
                }
            }
        }
    }
   static boolean compare(String s1, String s2) {
        boolean result = true;
        for(int i=0; i < s1.length && i < s2.length; i++) {
            if(s1.charAt(i) > s2.charAt(i)) {
                result = false;
                break;
            } else if(s1.charAt(i) < s2.chatAt(i)) {
                result = true;
                break;
            } else {
                if(s1.length() < s2.length()) {
                    result = true;
                } else {
                    result = false;
                }
            }
        }
        return result;
    }
}


【程序41】   题目:海滩上有一堆桃子,五只猴子来分。第一只猴子把这堆桃子平均分为五份,多了一个,这只猴子把多的一个扔入海中,拿走了一份。第二只猴子把剩下的桃子又平均分成五份,又多了一个,它同样把多的一个扔入海中,拿走了一份,第三、第四、第五只猴子都是这样做的,问海滩上原来最少有多少个桃子?

本题源码:
public class Demo41 {
    static int ts = 0;// 桃子总数
    int fs = 1;// 记录分的次数
    static int hs = 5;// 猴子数
    int tsscope = 5000;// 桃子数的取值范围,太大容易溢出
    public int fT(int t) {
        if (t == tsscope) {
            // 当桃子数到了最大的取值范围时取消递归
            System.out.println("结束");
            return 0;
        } else {
            if ((t - 1) % hs == 0 && fs <= hs) {
                if (fs == hs) {
                    System.out.println("桃子数=" + ts + "时满足分桃条件");
                }
                fs += 1;
                return fT((t - 1) / 5 * 4);// 返回猴子拿走一份后的剩下的总数
            } else {
                // 没满足条件
                fs = 1;// 分的次数重置为1
                return fT(ts += 1);// 桃子数加+1
            }
        }
    }

    public static void main(String[] args) {
           new Demo41().fT(0);
    }
}

个人修改:
public class Demo41 {
    public static void main(String[] args) {
        int sum = 0;
        for (int i = 6;; i++) {// 最少6个分最后一次
            sum = i;// 桃子数
            for (int j = 0; j < 5; j++) {// 分的次数循环
                if ((sum - 1) % 5 == 0 && j < 5) {// 如果扔一个后能均分5份,继续分
                    sum = (sum - 1) / 5 * 4;// 每分一次剩余桃子数
                        if (j == 4) {// 如果已分5次,且仍能除尽,输出,退出程序
                            System.out.println(i);
                            System.exit(0);
                        }
                }
            }
        }
    }
}


【程序42】   题目:809*??=800*??+9*??+1。其中??代表的两位数,8*??的结果为两位数,9*??的结果为3位数。求??代表的两位数,及809*??后的结果。
(本题为无解,去掉1有解)
public class Demo42 {
    public static void main(String[] args) {
        for (int i = 10; i < 100; i++) {
            if (809 * i == (800 * i + 9 * i + 1) && 8 * i >= 10 && 8 * i < 100
                && 9 * i >= 100 && 9 * i < 1000) {
                System.out.println("?? =" + i);
                System.out.println("809*??="+ 809 * i);
                System.exit(0);
            }
       }
    }
}


【程序43】   题目:求0—7所能组成的奇数个数。

暴力算法:
public class Demo43 {
    public static boolean isJiShu(int n) {
        if (n % 2 != 0) {
            return true;
        } else {
            return false;
        }
    }
    public static boolean fun(char c) {
        if (c >= '0' && c <= '7') {
            return true;
        } else {
            return false;
        }
    }
    public static void main(String[] args) {
        int count = 0;
        String s;
        for (int i = 0; i < 100000000; i++) {
            s = "" + i;
            boolean flag = true;
            char[] c = s.toCharArray();
            for (int j = 0; j < c.length; j++) {
                if (!fun(c[j])) {
                    flag = false;
                    break;
                }
            }
            if (flag && isJiShu(i)) {
                count++;
            }
            s = "";
        }
        System.out.println("共" + count + "个。");
    }
}

数学算法:
public class Demo43 {
    public static void main(String[] args) {
        // 因为是奇数,所以个位只能是1,3,5,7共4种,前面可随便排列
        int count = 4;// 个位的4种
        // 2位时,十位有8种,个位4种,8×4
        // 3位时,8×8×4……
        for (int i = 1; i < 8; i++) {
            count = 8 * count;
            System.out.println("count:" + count);
        }
    }
}

个人算法:
//组成1位数是4个
//组成2位数是7*4个
//组成3位数是7*8*4个
//组成4位数是7*8*8*4个
//......

public class Demo43 {
    public static void main (String[] args) {
        int sum=4;
        int j;
        System.out.println("组成1位数是 "+sum+" 个");
        sum=sum*7;
        System.out.println("组成2位数是 "+sum+" 个");
        for(j = 3; j <= 9; j++) {
            sum=sum*8;
            System.out.println("组成"+j+"位数是 "+sum+" 个");
        }
    }
}


【程序44】   题目:一个偶数总能表示为两个素数之和。(注:哥德巴赫猜想是想证明对任何大于6的自然数n之内的所有偶数可以表示为两个素数之和)

public class Demo44 {
    public static boolean isSuShu(int x) {
        if (x == 0 || x == 1) {
            return false;
        }
        for (int i = 2; i <= Math.sqrt(x); i++) {
            if (x % i == 0) {
                return false;
            }
        }
        return true;
    }

    public static void main(String[] args) {
        // 求了下100以内的情况
        for (int i = 0; i < 100; i = i + 2) {
            for (int j = 0; j <= (i + 1) / 2; j++) {
                if (isSuShu(j) && isSuShu(i - j)) {
                    System.out.println(i + "=" + j + "+" + (i - j));
                }
            }
        }
    }
}


public class Demo44{   
    public static void main(String[] args) {   
        for (int i=6;i<=100 ;i+=2 ) {   
            for (int j=2;j<100 ;j++ ) {       
                if (!isPrime(j) || !isPrime(i-j) || j >= i) {
                    continue;
                }
                System.out.println(i + "=" + j + "+" + (i-j));           
                break;       
            }       
        }   
    }   

    public static boolean isPrime(int n) {   
        for (int i = 2; i < n; i++) {
            if(n % i == 0) {
                return false;
            }   
        }           
        return true;   
    }
}


【程序45】   题目:(1)判断几个9能被一个素数整除。(2)判断一个整数能被几个9整除。(原题:一个素数能被几个9整除)

(一)
public class Demo45 {
    public static boolean isSuShu(int x) {
        if (x == 0 || x == 1) {
            return false;
        }
        for (int i = 2; i <= Math.sqrt(x); i++) {
            if (x % i == 0) {
                return false;
            }
        }
        return true;
    }

    public static void main(String[] args) {
        int[] a = new int[100];
        int n = 0;
        int num = 0;
        // 长度100的素数数组
        while (n < 100) {
            if (isSuShu(num)) {
                a[n] = num;
                n++;
                num++;
            } else {
                num++;
            }
        }
        /* for (int t : a) {
        System.out.println(t);
        }*/
        String s = "9";
        int index = 0;
        while (s.length() < 9) {
            if (new Integer(s).intValue() % a[index] == 0) {
                System.out.println(s + "%" + a[index] + "=0");
                if (index < 100 - 1) {
                    index++;
                } else {
                    index = 0;
                    s = s + "9";
                }
                // System.exit(0);
            } else {
                if (index < 100 - 1) {
                    index++;
                } else {
                index = 0;
                s = s + "9";
            }
        }
     }
  }
}

(二)
import java.util.*; public class Demo45 {
    public static void main (String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.print("请输入一个整数:");
        int num = in.nextInt();
        int tmp = num;
        int count = 0;
        for(int i = 0 ; tmp % 9 == 0 ;) {
            tmp = tmp / 9;
            count ++;
        }
        System.out.println(num+" 能够被 "+count+" 个9 整除。");
    }
}


【程序47】   题目:读取7个数(1—50)的整数值,每读取一个值,程序打印出该值个数的

import java.util.*;
public class Demo47 {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = 1,num;
        while(n <= 7) {
            do {
                System.out.print("请输入一个1--50 之间的整数:");
                num = s.nextInt();
            } while(num < 1 || num > 50);
            for(int i = 1;i <= num; i++) {
                System.out.print("*");
            }
            System.out.println();
            n ++;
        }
    }
}

import java.util.Scanner;
public class Demo47 {
    public static void print(int n) {
        for (int i = 0; i < n; i++) {
            System.out.print("*");
        }
        System.out.println();
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        for (int i = 0; i < 7; i++) {
            int temp = in.nextInt();
            print(temp);
        }
    }
}


【程序48】   题目:某个公司采用公用电话传递数据,数据是四位的整数,在传递过程中是加密的,加密规则如下:每位数字都加上5,然后用和除以10的余数代替该数字,再将第一位和第四位交换,第二位和第三位交换。

import java.util.Scanner;
public class Demo48{
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.println("请输入一个4位数字:");
        String str = in.next();
        if (!((str).matches("\\d{4}"))) {
            System.out.println("输入的不是4位数字!");
            System.exit(0);
        }
        char[] c = str.toCharArray();
        int[] a = new int[4];
        for (int i = 0; i < a.length; i++) {
            a[i] = ((int) (c[i] - '0') + 5) % 10;
        }
        int t;
        t = a[0];
        a[0] = a[3];
        a[3] = t;
        t = a[1];
        a[1] = a[2];
        a[2] = t;
        System.out.println("结果是:" + a[0] + a[1] + a[2] + a[3]);
    }
}

import java.util.*;
public class Demo48 {
    public static void main(String args[]) {
        Scanner s = new Scanner(System.in);
        int num=0,temp;
        do {
            System.out.print("请输入一个4位正整数:");
            num = s.nextInt();
        } while (num < 1000 || num > 9999);
            int a[] = new int[4];
            a[0] = num/1000; //取千位的数字
            a[1] = (num/100) % 10; //取百位的数字
            a[2] = (num/10) % 10; //取十位的数字
            a[3] = num % 10; //取个位的数字
            for(int j = 0;j < 4; j++) {
                a[j] += 5;
                a[j] %= 10;
            }
            for(int j= 0; j<= 1; j++) {
                temp = a[j]; a[j] = a[3-j]; a[3-j] = temp;
            }
            System.out.print("加密后的数字为:");
            for(int j = 0; j < 4; j++) {
                System.out.print(a[j]);
            }
      }
}


【程序49】   题目:计算字符串中子串出现的次数。

import java.util.Scanner;
public class Demo49 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.println("请输入主串:");
        String str1 = in.nextLine();
        System.out.println("请输入子串:");
        String str2 = in.nextLine();
        // 生成子串长度的N个字符串数组
        String[] sa = new String[str1.length() - str2.length() + 1];
        for (int i = 0; i < sa.length; i++) {
            sa[i] = str1.substring(i, i + str2.length());
        }
        int sum = 0;
        // 子串与N个拆开的子串比对
        for (int i = 0; i < sa.length; i++) {
            if (sa[i].equals(str2)) {
                // 成功配对,计数器+1;
                sum++;
                // 因为不计算重叠的子串,所以跳过配对之后的部分拆分子串
                i = i + str2.length();
            }
        }
        System.out.println("主串中共包含" + sum + "个字串");
    }
}

【程序50】    题目:有五个学生,每个学生有3门课的成绩,从键盘输入以上数据(包括学生号,姓名,三门课成绩),计算出平均成绩,把原有的数据和计算出的平均分数存放在磁盘文件

import java.io.File;
import java.io.FileWriter;
import java.util.Scanner;
class Student {
    private int number = 0;
    private String name = "";
    private double[] a = new double[3];
    public double getAve() {
        return (a[0] + a[1] + a[2]) / 3;
    }

    public Student(int number, String name, double[] a) {
        super();
        this.number = number;
        this.name = name;
        this.a = a;
    }

    @Override
    public String toString() {
        return "学号:" + this.number + "\t姓名:" + this.name + "\r\n各科成绩:\r\n" + a[0] + "\t" + a[1] + "\t" + a[2] + "\r\n平均成绩:\r\n" + this.getAve();
    }
}

public class Demo50 {
    public static Student input() {
        Scanner s = new Scanner(System.in);
        System.out.println("请输入学号:");
        int num = s.nextInt();
        System.out.println("请输入姓名:");
        String name = s.next();
        System.out.println("请分别输入3门成绩:");
        double[] a = new double[3];
        for (int i = 0; i < 3; i++) {
            a[i] = s.nextDouble();
        }
        return new Student(num, name, a);
    }

    public static void main(String[] args) throws Exception {
        Student[] st = new Student[2];
        for (int i = 0; i < st.length; i++) {
            st[i] = input();
        }
        File f = new File("d:" + File.separator + "123.txt");
        FileWriter output = new FileWriter(f);
        for (int i = 0; i < st.length; i++) {
            output.write(st[i].toString() + "\r\n");
            output.write("\r\n");
        }
        output.close();
    }
}

最后编辑于
©著作权归作者所有,转载或内容合作请联系作者
  • 序言:七十年代末,一起剥皮案震惊了整个滨河市,随后出现的几起案子,更是在滨河造成了极大的恐慌,老刑警刘岩,带你破解...
    沈念sama阅读 158,736评论 4 362
  • 序言:滨河连续发生了三起死亡事件,死亡现场离奇诡异,居然都是意外死亡,警方通过查阅死者的电脑和手机,发现死者居然都...
    沈念sama阅读 67,167评论 1 291
  • 文/潘晓璐 我一进店门,熙熙楼的掌柜王于贵愁眉苦脸地迎上来,“玉大人,你说我怎么就摊上这事。” “怎么了?”我有些...
    开封第一讲书人阅读 108,442评论 0 243
  • 文/不坏的土叔 我叫张陵,是天一观的道长。 经常有香客问我,道长,这世上最难降的妖魔是什么? 我笑而不...
    开封第一讲书人阅读 43,902评论 0 204
  • 正文 为了忘掉前任,我火速办了婚礼,结果婚礼上,老公的妹妹穿的比我还像新娘。我一直安慰自己,他们只是感情好,可当我...
    茶点故事阅读 52,302评论 3 287
  • 文/花漫 我一把揭开白布。 她就那样静静地躺着,像睡着了一般。 火红的嫁衣衬着肌肤如雪。 梳的纹丝不乱的头发上,一...
    开封第一讲书人阅读 40,573评论 1 216
  • 那天,我揣着相机与录音,去河边找鬼。 笑死,一个胖子当着我的面吹牛,可吹牛的内容都是我干的。 我是一名探鬼主播,决...
    沈念sama阅读 31,847评论 2 312
  • 文/苍兰香墨 我猛地睁开眼,长吁一口气:“原来是场噩梦啊……” “哼!你这毒妇竟也来了?” 一声冷哼从身侧响起,我...
    开封第一讲书人阅读 30,562评论 0 197
  • 序言:老挝万荣一对情侣失踪,失踪者是张志新(化名)和其女友刘颖,没想到半个月后,有当地人在树林里发现了一具尸体,经...
    沈念sama阅读 34,260评论 1 241
  • 正文 独居荒郊野岭守林人离奇死亡,尸身上长有42处带血的脓包…… 初始之章·张勋 以下内容为张勋视角 年9月15日...
    茶点故事阅读 30,531评论 2 245
  • 正文 我和宋清朗相恋三年,在试婚纱的时候发现自己被绿了。 大学时的朋友给我发了我未婚夫和他白月光在一起吃饭的照片。...
    茶点故事阅读 32,021评论 1 258
  • 序言:一个原本活蹦乱跳的男人离奇死亡,死状恐怖,灵堂内的尸体忽然破棺而出,到底是诈尸还是另有隐情,我是刑警宁泽,带...
    沈念sama阅读 28,367评论 2 253
  • 正文 年R本政府宣布,位于F岛的核电站,受9级特大地震影响,放射性物质发生泄漏。R本人自食恶果不足惜,却给世界环境...
    茶点故事阅读 33,016评论 3 235
  • 文/蒙蒙 一、第九天 我趴在偏房一处隐蔽的房顶上张望。 院中可真热闹,春花似锦、人声如沸。这庄子的主人今日做“春日...
    开封第一讲书人阅读 26,068评论 0 8
  • 文/苍兰香墨 我抬头看了看天上的太阳。三九已至,却和暖如春,着一层夹袄步出监牢的瞬间,已是汗流浃背。 一阵脚步声响...
    开封第一讲书人阅读 26,827评论 0 194
  • 我被黑心中介骗来泰国打工, 没想到刚下飞机就差点儿被人妖公主榨干…… 1. 我叫王不留,地道东北人。 一个月前我还...
    沈念sama阅读 35,610评论 2 274
  • 正文 我出身青楼,却偏偏与公主长得像,于是被迫代替她去往敌国和亲。 传闻我的和亲对象是个残疾皇子,可洞房花烛夜当晚...
    茶点故事阅读 35,514评论 2 269

推荐阅读更多精彩内容

  • 【程序1】 题目:古典问题:有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子长到第三个月后每个月又生一...
    阿里高级软件架构师阅读 3,245评论 0 19
  • 【程序1】 题目:古典问题:有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子长到第三个月后每个月又生一对兔...
    叶总韩阅读 5,066评论 0 41
  • Java经典问题算法大全 /*【程序1】 题目:古典问题:有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子...
    赵宇_阿特奇阅读 1,789评论 0 2
  • 50道经典Java编程练习题,将数学思维运用到编程中来。抱歉哈找不到文章的原贴了,有冒犯的麻烦知会声哈~ 1.指数...
    OSET我要编程阅读 6,693评论 0 9
  • 昨天,范伟摘得金马影帝让很多人感到意外,但是细想想又觉得实至名归。虽然他和小李子获奥斯卡影帝的状况不太一样,但是没...
    彼得猫乌金阅读 241评论 0 1