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题目：
Given a list of integers (A1, A2, ..., An), 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, ..., An(1 <= Ai <= 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
解题报告：确实刚上来是真的看不懂题目是意思，求解小于等于m的数能被数组中的n个数整除的数目和。打算直接遍历走完是不现实的，因为m的范围太大了， 所以想到了容斥定理，数组中的每个数字能被整除的数目的求并，自然引出容斥定理，然后套模板，最后做差输出即可。
ac代码：
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int num[15],n,m;
int gcd(int a,int b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
} //最大公因子
int lcm(int a,int b)
{
return a/gcd(a,b)*b;
}//最小公倍数
int dfs(int pos,int val)
{
if(pos==n)
{
return m/val;
}
return dfs(pos+1,val)-dfs(pos+1,lcm(val,num[pos]));
}
//容斥定理
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=0;i<n;i++)
scanf("%d",&num[i]);
printf("%d\n",m-dfs(0,1));
}
return 0;
}