题目描述：12个金币中有一个是假币，假币的大小颜色都和真的一样，只有重量不一样，但是有可能比真的轻，也有可能重，现在要通过称重找出哪个假的。
输入格式：
1
ABCD EFGH even
ABCI EFJK up
ABIJ EFGH even
说明：第三个字符串有三种情况”even”,”up”,”down”.分别表示平衡，左边轻，右边轻。
输出格式：
K is the counterfeit coin and it is light.
思路：先用穷举将所有的表达式过滤，两边都相等的肯定是真币就是true，不为true的就判断被怀疑的次数，被怀疑的次数最多的，就是假币。
java实现代码：
import java.util.Scanner;
public class TestForMySon {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
if (scan.hasNext()) {
int n = Integer.parseInt(scan.nextLine().trim());
for (int i = 0; i < n; i++) {
boolean[] real = {false, false, false, false, false, false, false, false, false, false, false, false};
int[] hy = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
String[] s1 = scan.nextLine().trim().split(" ");
String[] s2 = scan.nextLine().trim().split(" ");
String[] s3 = scan.nextLine().trim().split(" ");
char[] c11 = s1[0].toCharArray();
char[] c12 = s1[1].toCharArray();
char[] c21 = s2[0].toCharArray();
char[] c22 = s2[1].toCharArray();
char[] c31 = s3[0].toCharArray();
char[] c32 = s3[1].toCharArray();
if (s1[2].equals("even")) {
for (int j = 0; j < c11.length; j++) {
real[c11[j] - 'A'] = true;
}
for (int j = 0; j < c12.length; j++) {
real[c12[j] - 'A'] = true;
}
}
if (s2[2].equals("even")) {
for (int j = 0; j < c21.length; j++) {
real[c21[j] - 'A'] = true;
}
for (int j = 0; j < c22.length; j++) {
real[c22[j] - 'A'] = true;
}
}
if (s3[2].equals("even")) {
for (int j = 0; j < c31.length; j++) {
real[c31[j] - 'A'] = true;
}
for (int j = 0; j < c32.length; j++) {
real[c32[j] - 'A'] = true;
}
}
if (s1[2].equals("up")) {
for (int j = 0; j < c11.length; j++) {
hy[c11[j] - 'A']++;
}
for (int j = 0; j < c12.length; j++) {
hy[c12[j] - 'A']--;
}
}
if (s2[2].equals("up")) {
for (int j = 0; j < c21.length; j++) {
hy[c21[j] - 'A']++;
}
for (int j = 0; j < c22.length; j++) {
hy[c22[j] - 'A']--;
}
}
if (s3[2].equals("up")) {
for (int j = 0; j < c31.length; j++) {
hy[c31[j] - 'A']++;
}
for (int j = 0; j < c32.length; j++) {
hy[c32[j] - 'A']--;
}
}
if (s1[2].equals("down")) {
for (int j = 0; j < c11.length; j++) {
hy[c11[j] - 'A']--;
}
for (int j = 0; j < c12.length; j++) {
hy[c12[j] - 'A']++;
}
}
if (s2[2].equals("down")) {
for (int j = 0; j < c21.length; j++) {
hy[c21[j] - 'A']--;
}
for (int j = 0; j < c22.length; j++) {
hy[c22[j] - 'A']++;
}
}
if (s3[2].equals("down")) {
for (int j = 0; j < c31.length; j++) {
hy[c31[j] - 'A']--;
}
for (int j = 0; j < c32.length; j++) {
hy[c32[j] - 'A']++;
}
}
int m = 0;
for (int k = 0; k < real.length; k++) {
if (!real[k]) {
int max = Math.abs(hy[k]);
m = k;
for (int j = k + 1; j < hy.length; j++) {
if (Math.abs(hy[j]) > max && !real[j]) {
max = Math.abs(hy[j]);
m = j;
}
}
break;
}
}
char key = (char) ('A' + m);
if (hy[m] > 0) {
System.out.println(key + " is the counterfeit coin and it is heavy.");
} else {
System.out.println(key + " is the counterfeit coin and it is light.");
}
}
}
}
}
