PAT Advanced 1023. Have Fun with Numbers (20) (C语言实现)

我的PAT系列文章更新重心已移至Github，欢迎来看PAT题解的小伙伴请到Github Pages浏览最新内容(本篇文章链接)。此处文章目前已更新至与Github Pages同步。欢迎star我的repo。

题目

Notice that the number 123456789 is a 9-digit number consisting exactly the
numbers from 1 to 9, with no duplication. Double it we will obtain 246913578,
which happens to be another 9-digit number consisting exactly the numbers from
1 to 9, only in a different permutation. Check to see the result if we double
it again!

Now you are suppose to check if there are more numbers with this property.
That is, double a given number with $k$ digits, you are to tell if the
resulting number consists of only a permutation of the digits in the original
number.

Input Specification:

Each input contains one test case. Each case contains one positive integer
with no more than 20 digits.

Output Specification:

For each test case, first print in a line "Yes" if doubling the input number
gives a number that consists of only a permutation of the digits in the
original number, or "No" if not. Then in the next line, print the doubled
number.

Sample Input:

1234567899

Sample Output:

Yes
2469135798

思路

主要是大数运算，这里是大数和个位数的乘法，我的简要思路是：

用字符数组倒序存储数字，即小序数的地方存低位数字。
做乘法，模拟手算，注意字符和数字的转换：
- s代表算出的当前位，
- l代表当前剩余的进位数字，在下一次计算时，加上此数字。
将结果再翻转回来，便于输出。
统计两个数包含数字的数量之差，按要求输出即可。

代码

最新代码@github，欢迎交流

#include <stdio.h>
#include <string.h>

void reverse(char n[])
{
    char temp;
    int len = strlen(n);
    for(int i = 0; i < len / 2; i++)
    {
        temp = n[i];
        n[i] = n[len - i - 1];
        n[len - i - 1] = temp;
    }
}

void doubleLargeNumber(char n[], char n2[])
{
    int l, s = 0, len = strlen(n);

    for(int i = 0; i < len; i++)
    {
        s += 2 * (n[i] - '0');
        l = s / 10;
        s %= 10;
        n2[i] = s + '0';

        s = l;
    }

    if(s)
        n2[len] = s + '0';
}

int main()
{
    char N[22] = {0}, N2[22] = {0};
    int diffcount[10] = {0};

    scanf("%s", N);
    /* store number in reverse order */
    reverse(N);
    doubleLargeNumber(N, N2);
    /* reverse back (just need N2) for printing */
    reverse(N2);

    /* Count the difference of counts of digits 0-9 */
    for(int i = 0; i < strlen(N); i++)
        diffcount[N[i] - '0']++;
    for(int i = 0; i < strlen(N2); i++)
        diffcount[N2[i] - '0']--;

    /* Check */
    for(int i = 0; i < 10; i++)
        if(diffcount[i])
        {
            printf("No\n%s", N2);
            return 0;
        }
    printf("Yes\n%s", N2);

    return 0;
}