zoj 2836 Number Puzzle（容斥，求1到m有多少个数能被数组里任意一个整除）

Given a list of integers (A₁, A₂, ..., A_n), and a positive integer M, please find the number of positive integers that are not greater than M and dividable by any integer from the given list.

Input

The input contains several test cases.

For each test case, there are two lines. The first line contains N (1 <= N <= 10) and M (1 <= M <= 200000000), and the second line contains A₁, A₂, ..., A_n(1 <= A_i <= 10, for i = 1, 2, ..., N).

Output

For each test case in the input, output the result in a single line.

Sample Input

3 2
2 3 7
3 6
2 3 7

Sample Output

1

4

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1836

#include<iostream>#include<algorithm>#include<string>#include<map>#include<vector>#include<cmath>#include<string.h>#include<stdlib.h>#include<cstdio>#define ll long longusing namespace std;ll n,m,s;int x[15];ll gcd(ll a,ll b){        return b?gcd(b,a%b):a;}void dfs(ll aim,ll step,ll k,ll per,ll u){//k:从第k个开始 per:当前数的最小公倍数 if(aim==step){if(u==1)s+=m/per; //m除所有数的最小公倍数 elses-=m/per;return;}elsefor(int i=k;i<n;++i) //这个K真是妙极了！（保证了每一种组合的存在） dfs(aim,step+1,i+1,(per*x[i])/gcd(per,x[i]),u);}int main(){while(cin>>n>>m){s=0;for(int i=0;i<n;++i)cin>>x[i];for(int i=1;i<=n;++i){if(i%2==1)dfs(i,0,0,1,1);//1个数的时候加，2个减，3个加，4个减... elsedfs(i,0,0,1,0);}cout<<s<<endl;}return 0;}