容斥定理

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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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <string>
#include <deque>
#include <map>
#include <vector>
#include <stack>
using namespace std;
#define  ll long long 
#define  N   29
#define  M 1000000000
#define  gep(i,a,b)  for(int  i=a;i<=b;i++)
#define  gepp(i,a,b) for(int  i=a;i>=b;i--)
#define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
#define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
#define  mem(a,b)  memset(a,b,sizeof(a))
#define  ph  push_back
/*
int、long long类型都可以，需要注意的是两个类型必须要相同
还有就是不能用浮点型，当然也可以手写gcd函数，它头文件是algorithm。
*/

ll lcm(ll a,ll b){
    ll c=__gcd(a,b);
    return a/c*b;
}
int t;
ll n,m,a[N];
//C(n,1)+C(n,2)+C(n,3)+……+C(n,n)==(2^n)-1
int main()
{
    scanf("%d",&t);
    while(t--){
        scanf("%lld%lld",&n,&m);
        ll sum=0;
        gep(i,0,n-1) scanf("%lld",&a[i]);
        gep(i,1,(1<<n)-1){//最大(1<<n)-1，C(n,n)
            ll cnt=0,ans=1;
            gep(j,0,n-1){//j从0开始
                 if(i&(1<<j)){
                     cnt++;//C(n,cnt)
                     ans=lcm(ans,a[j]);
                     if(ans>m) break;
                 }                  
            }
            sum+=cnt%2?m/ans:-m/ans;//容斥定理+1-2+3-4…… 
        }
        printf("%lld
",sum);
    }
    return 0;
}

如果被计数的事物有A、B、C三类，那么，A类和B类和C类元素个数总和= A类元素个数+ B类元素个数+C类元素个数—既是A类又是B类的元素个数—既是A类又是C类的元素个数—既是B类又是C类的元素个数+既是A类又是B类而且是C类的元素个数。
（A∪B∪C = A+B+C - A∩B - B∩C - C∩A + A∩B∩C）