AtCoder ABC423F - Loud Cicada 题解 容斥原理

题目大意：

问 \([1, Y]\) 范围内有多少个数是：\(A_1, A_2, \ldots, A_N\) 中恰好 \(K\) 个数的倍数。

解题思路：

容斥原理。

思路完全来自 StelaYuri大佬的博客。

示例程序：

#include <bits/stdc++.h>
using namespace std;void write(__int128 a) {if (a / 10)write(a / 10);cout << (int) (a % 10);
}
__int128 gcd(__int128 a, __int128 b) {if (!b) return a;return gcd(b, a%b);
}
__int128 lcm(__int128 a, __int128 b) {return a / gcd(a, b) * b;
}int n, m;
long long Y, a[22];__int128 C[22][22];void init() {for (int i = 0; i <= 20; i++) {C[i][0] = C[i][i] = 1;for (int j = 1; j < i; j++)C[i][j] = C[i-1][j-1] + C[i-1][j];}
}int main() {init();cin >> n >> m >> Y;for (int i = 0; i < n; i++)cin >> a[i];__int128 ans = 0;for (int s = 1; s < (1<<n); s++) {int k = __builtin_popcount(s);if (k < m)continue;__int128 D = 1;for (int i = 0; i < n; i++) {if ((s >> i) & 1) {D = lcm(D, a[i]);if (D > Y)break;}}if (D > Y) continue;__int128 tmp = C[k][m] * (Y / D);if ((k - m) % 2)ans -= tmp;elseans += tmp;}write(ans);return 0;
}