How many integers can you find（容斥原理）

How many integers can you find

   Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7. 

Input
There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. and the M integer are non-negative and won’t exceed 20.
Output
For each case, output the number.
Sample Input

12 2
2 3

Sample Output

7

题意：

给定n和一个大小为m的集合，集合元素为非负整数。为1…n内能被集合里任意一个数整除的数字个数。n<=2^31,m<=10

分析：

首先我们将能被i整除记为满足性质$A_i$，那么根据题意实际是让我们求

$|A_1\bigcup A_2 \bigcup \cdots \bigcup A_m|$

根据容斥原理的推论：

$|A_1\bigcup A_2 \bigcup \cdots \bigcup A_m|$

$=\sum|A_i|-\sum|A_i\bigcup A_j| + \sum |A_i\bigcup A_j\bigcup A_k| +\cdots$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +(-1)^{m+1}|A_1\bigcup A_2 \bigcup \cdots \bigcup A_m|$

根据组合数我们知道它的项数一共有：

$C_m^0+C_m^1+C_m^2 + \cdots + C_m^m = 2^m$

因为m不超过10，所以完全可以枚举完成

巧妙利用二进制表示取法1代表取0表示不取，然后用按位与得到选取了哪一个

满足奇数个性质是加号，满足偶数个性质是减号

code：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,m,a[20];
ll gcd(ll a,ll b){
    return b ? gcd(b,a%b) : a;
}
int main(){
    while(scanf("%lld%lld",&n,&m) != EOF){
        int ans = 0;
        vector<int> v;
        for(int i = 0; i < m; i++){
            scanf("%lld",&a[i]);
            if(a[i] > 0) v.push_back(a[i]);//注意不能有0
        }
        m = v.size();//更新长度
        for(int i = 1; i < (1 << m); i++){//注意这里其实枚举到(1<<m)-1即m个1表示m个性质全有
            int cnt = 0;
            ll x = 1;
            for(int t = 0; t < m; t++){//枚举每个性质看当前情况选择了那个性质
                if(i & (1 << t)){
                    cnt++;
                    x = x * v[t] / gcd(x,v[t]);
                }
            }
            if(cnt & 1) ans += (n - 1) / x;
            else ans -= (n - 1) / x;
        }
        printf("%d\n",ans);
    }
    return 0;
}