题意：在给出的16个数中，求使得满足
x1* 4 +
x2* 3 + x3* 2 + x4 =
x5 + x6* 2 + x7* 3 +
x8* 4
y1* 4 +
y2* 3 + y3* 2 + y4 =
y5 + y6* 2 + y7* 3 +
y8* 4
这两个等式的个数有多少。
思路：状态枚举，先枚举四个数，然后查找是否有与这四个数组成的等式和的结果一样的组合，且这个组合与这四个数不相冲突，最后就是通过计算
st[i]*st[i^state]的总和就是，state表示全都有的状态
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
int a[20],num[20];
vector<int> val[12000];
int st[1<<17];
bool judge(int x){
int cnt = 0;
for (int i = 0; i < 16; i++){
if (x & (1<<i))
num[cnt++] = a[i];
}
return cnt == 4;
}
int main(){
int t = 0;
while (scanf("%d",&a[0]) != EOF && a[0]){
for (int i = 1; i < 16; i++)
scanf("%d",&a[i]);
for (int i = 0; i < 12000; i++)
val[i].clear();
memset(st,0,sizeof(st));
sort(a,a+16);
for (int i = 0; i < (1<<17); i++){
if (judge(i)){
do{
int temp = num[0]*4+num[1]*3+num[2]*2+num[3];
for (int j = 0; j < val[temp].size(); j++)
if ((i & val[temp][j]) == 0)
st[i|val[temp][j]]++;
val[temp].push_back(i);
}
while (next_permutation(num,num+4));
}
}
long long ans = 0;
for (int i = 0; i < (1<<16); i++)
ans += st[i] * st[i^((1<<16)-1)];
printf("Case %d: %lld\n",++t,ans/2);
}
return 0;
}