POJ 3904 Sky Code 容斥原理

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题目来源：POJ 3904 Sky Code

题意：选出最大公约数为1的四元组的方案

思路：容斥原理总的方案C(n,4)减去t(1)+t(2)-t(3)+...+(-)^kt(k)

t(i)表示四元组质因子的个数为i的方案数

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 10010;typedef long long LL;int a[maxn];int b[maxn];int cnt[maxn];int n, m;//返回a^p mod n 快速幂bool vis[maxn];int prime[maxn]; LL C(LL n, LL m){if(n < m)return 0;LL ans = 1;for(int i = 1; i <= m; i++){ans *= n--;ans /= i;}return ans;}int main(){while(scanf("%d", &n) != EOF){memset(a, 0, sizeof(a));for(int i = 0; i < n; i++){scanf("%d", &b[i]);int x = b[i];int sum = 0;for(int j = 2; j*j <= x; j++){if(x%j == 0){prime[sum++] = j;x /= j;while(x%j == 0)x /= j;}}if(x > 1)prime[sum++] = x;for(int j = 1; j < (1<<sum); j++){int temp = 1;int num = 0;for(int k = 0; k < sum; k++){if(j&(1<<k)){temp *= prime[k];num++;}}a[temp]++;cnt[temp] = num;}}LL ans = 0;for(int i = 0; i <= 10000; i++){if(!a[i])continue;if(cnt[i]&1)ans += C(a[i], 4);elseans -= C(a[i], 4);}printf("%I64d\n", C(n, 4)-ans);}return 0;}

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