排斥定理

Teemo's dream

Teemo decides to use his money to conquer the universe.

It is known that there are m planets that humans can reach at present. They are numbered from 1 to m. Teemo bought n kinds of gateways. Their IDs are a1, a2, ..., an, the gateway whose ID is ai can transmit Teemo to the stars numbered ai,2ai, 3ai, ..., k*ai (1<=k*ai<=m, k is a positive integer), now Teemo wants to know, how many planets can he reach?

 

Input Format

On the firstline one positive number: the number of test cases, at most 20. After that per test case:

• One line contains two integers n and m, (1 <= n <= 15, 1<= m < = 1e9), respectively represent the number of the gateway, the number of the stars that humans can reach.

• One line contains n integers, the i-th integer a[i], indicating that the ID of the  i-th gateway is a[i], (2<=a[i]<=1e9).

 

Ouput Format

Per test case:

• One line contains an integer, which indicates how many planets Teemo can reach at most.

 

样例输入

2
2 15
2 3
5 100
2 3 4 5 6

样例输出

10
74

 

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int t, n;
ll m, a[20];
int main() {
    cin >> t;
    while(t--) {
        cin >> n >> m;
        for(int i = 0; i < n; i ++) cin >> a[i];
        ll ans = 0;
        for(int state = 1; state < (1<<n); ++state) {
            ll cnt = 0, _lcm = 1;
            for(int i = 0; i < n; i ++) {
                if(state&(1<<i)) {
                    _lcm = _lcm/__gcd(a[i],_lcm)*a[i];
                    cnt ++;
                    if(_lcm > m) break;
                }
            }
            ans += cnt%2 ? m/_lcm: -m/_lcm;
        }
        cout << ans << endl;
    }
    return 0;
}