POJ 3139 Balancing the Scale

Description

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:

x₁ * 4+x₂ * 3+x₃ * 2+x₄=x₅+x₆ * 2+x₇ * 3+x₈ * 4y₁ * 4+y₂ * 3+y₃ * 2+y₄=y₅+y₆ * 2+y₇ * 3+y₈ * 4

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 85 880

Sample Output

Case 1: 15227223

算有多组答案，先根据数值的范围可以知道，最大值不会超过10240。

先用四个数组成全部的情况，总共有4!*C(16,4)种，用链表分别对应到相应的值上。

然后对于同一个值，计算出对应的8个数的组合，用状压统计到数组里

然后两个互补的8个数组合相乘就是答案。

#include<set>#include<map>#include<ctime>#include<cmath>#include<stack>#include<queue>#include<bitset>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define rep(i,j,k) for (int i = j; i <= k; i++)#define per(i,j,k) for (int i = j; i >= k; i--)#define loop(i,j,k) for (int i = j;i != -1; i = k[i])#define lson x << 1, l, mid#define rson x << 1 | 1, mid + 1, r#define fi first#define se second#define mp(i,j) make_pair(i,j)#define pii pair<int,int>using namespace std;typedef long long LL;const int low(int x) { return x&-x; }const double eps = 1e-4;const int INF = 0x7FFFFFFF;const int mod = 9973;const int N = 1 << 16;int a[16], b[4], c[4], cas = 0;int ft[N], nt[N], u[N], sz;int f[N];void get(int x, int y, int z){if (x == 4){rep(i, 0, 3) c[i] = b[i];do{int C = 0;rep(i, 0, 3) C += a[c[i]] * (4 - i);u[sz] = z; nt[sz] = ft[C]; ft[C] = sz++;} while (next_permutation(c, c + 4));}else rep(i, y, 15){b[x] = i;get(x + 1, i + 1, z | (1 << i));}}int main(){while (scanf("%d", &a[0]) != EOF, a[0]){rep(i, 1, 15) scanf("%d", &a[i]);rep(i, 0, N) ft[i] = -1;sz = 0; get(0, 0, 0);memset(f, 0, sizeof(f));rep(i, 0, N){loop(j, ft[i], nt){loop(k, nt[j], nt){if (u[j] & u[k]) continue;f[u[j] | u[k]]++;}}}LL ans = 0;rep(i, 0, (1 << 16) - 1){ans += 1LL * f[i] * f[((1 << 16) - 1) ^ i];}printf("Case %d: %lld\n", ++cas, ans / 2); }return 0;}