HDU 1796 How many integers can you find（组合数学-容斥原理）

How many integers can you find

Problem Description

  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. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.

 

Output

  For each case, output the number.

 

Sample Input

12 22 3

 

Sample Output

7

 

Author

wangye

 

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

 

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题目大意：

给你1个数n，再给m个数，问你1~n-1里面有多少个数能被这m个数的任意一个数整除。

解题思路：

利用容斥原理就可以解决。

解题代码：

#include <iostream>#include <cstdio>#include <vector>using namespace std;typedef long long ll;int n,m,a[20];ll gcd(ll a,ll b){    return b>0 ? gcd(b,a%b):a;}int main(){    while(scanf("%d%d",&n,&m)!=EOF){        int ans=0;        vector <int> v;        for(int i=0;i<m;i++){            scanf("%d",&a[i]);            if(a[i]>0) v.push_back(a[i]);        }        m=v.size();        for(int i=1;i<(1<<m);i++){            int cnt=0;            ll x=1;            for(int t=0;t<m;t++){                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;        }        cout<<ans<<endl;    }    return 0;}