UVA

题目：

代码：

#include<iostream>using namespace std;int gcd(int a, int b){if (b)return gcd(b, a%b);return a;}int main(){int n, m, l[15], s;while (cin >> n >> m){s = 0;for (int i = 0; i < m; i++)cin >> l[i];for (int k = 0; k < (1 << m); k++){long long r = 1, b = 0;for (int i = 0; i < m; i++)if (k&(1 << i)){if (r <= n)r *= l[i] / gcd(l[i], r);b++;}if (b % 2)s -= n/r;else s += n/r;}cout << s << endl;}return 0;}

用了状态压缩来枚举2^m种状态，用k对应状态

注意一个细节，当r超过n时，r/n就已经是0了，就不要再继续求lcm了，否则会越界，因为多个数的lcm很容易超过long long的范围