You are given a strange scale (see the figure below), and you are wondering how to balance this scale. After several attempts, you have discovered the way to balance it -- you need to put different numbers on different squares while satisfying the following two equations:
How many ways can you balance this strange scale with the given numbers?
Input
There are multiple test cases in the input file. Each test case consists of 16 distinct numbers in the range [1, 1024] on one separate line. You are allowed to use each number only once.
A line with one single integer 0 indicates the end of input and should not be processed by your program.
Output
For each test case, if it is possible to balance the scale in question, output one number, the number of different ways to balance this scale, in the format as indicated in the sample output. Rotations and reversals of the same arrangement should be counted only once.
Sample Input
87 33 98 83 67 97 44 72 91 78 46 49 64 59 85 88 0
Sample Output
Case 1: 15227223
题意：给定16个数字，求题目中给定公式的平衡方法有几种
思路：技巧枚举+位运算，通过位运算每次枚举4个数字，通过全排列求出和，算出2边相同时候状态的种数，最后在把x的情况和y的情况数相乘并所有都相加就是答案，计算全排列前要排序一开始忘了答案一直少，后面看了题解才发现。一开始还理解错题目了以为x和y的和都是要相同的。
代码：
#include <stdio.h>
#include <string.h>
#include <vector>
#include <algorithm>
using namespace std;
int num[16], st[16], Sum[(1<<16)];
vector<int> sum[22222];
bool judge(int state) {
int sn = 0;
for (int i = 0; i < 16; i++) {
if (state&(1<<i))
st[sn++] = num[i];
}
if (sn == 4) return true;
return false;
}
int solve() {
int ans = 0;
memset(sum, 0, sizeof(sum));
memset(Sum, 0, sizeof(Sum));
sort(num, num + 16);
for (int i = 0; i < (1<<16); i++) {
if (judge(i)) {
do {
int s = st[0] * 4 + st[1] * 3 + st[2] * 2 + st[3];
for (int j = 0; j < sum[s].size(); j++) {
if ((i&sum[s][j]) == 0)
Sum[(i|sum[s][j])]++;
}
sum[s].push_back(i);
} while(next_permutation(st, st + 4));
}
}
for (int k = 0; k < (1<<16); k++)
ans += Sum[k] * Sum[k^((1<<16)-1)];
return ans / 2;
}
int main() {
int cas = 0;
while (~scanf("%d", &num[0]) && num[0]) {
for (int i = 1; i < 16; i++)
scanf("%d", &num[i]);
printf("Case %d: %d\n", ++cas, solve());
}
return 0;
}