UVA 10325 The Lottery

容斥原理求集合并集.

这里的集合就是给出的M个数,每个数的倍数(小于等于N)就是一个集合.

先求出最小公倍数,然后 N/最小公倍数 就可以得出每个子集的元素数量.

#include <iostream>#include <cstdio>using namespace std;long long gcd(long long a, long long b){return b == 0 ? a : gcd(b, a % b);}long long lcm(long long a, long long b){return a / gcd(a, b) * b;}long long N, M, sel[16];int main(){while (scanf("%lld%lld", &N, &M) == 2){long long ans = N;long long uni = 0;for (int i = 0; i < M; ++i){scanf("%lld", &sel[i]);}for (long long i = 0; i < (1 << M); ++i){long long cnt = 0, lcmn = 1;for (long long j = 0; j < M; ++j){if ((1 << j) & i){cnt++;lcmn = lcm(lcmn, sel[j]);}}if(cnt == 0)continue;if(cnt & 1){uni += N / lcmn;}else{uni -= N / lcmn;}}printf("%d\n", ans - uni);}return 0;}