HDU 1796 How many integers can you find 容斥入门

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 2 2 3

 

Sample Output

7

 

题意：

　　　给你m个数的集合，给你一个n问你小于n并且是这m个数里面的数的倍数有多少

题解：

　　　dfs容斥原理

　　　奇数偶数加减法

#include<bits/stdc++.h>
using namespace std;
const int N = 3e6+20, M = 1e6+10, mod = 1e9+7,inf = 1e9+1;

typedef long long ll;

ll ans;
int x,a[N],n,m;
ll gcd(ll a,ll b) {return b==0?a:gcd(b,a%b);}
void dfs(int i,int num,ll tmp) {
     if(i>=m) {
        if(num==0) ans=0;
        else {
            if(num&1) ans = (ans+n/tmp);
            else ans = ans - n/tmp;
        }
        return ;
     }
     dfs(i+1,num,tmp);
     dfs(i+1,num+1,tmp*a[i]/gcd(tmp,a[i]));
}
int main() {
    while(scanf("%d%d",&n,&m)!=EOF) {
        int k = 0;
        n--;
        for(int i=1;i<=m;i++) {
            scanf("%d",&x);
            if(x) a[k++] = x;
        }
        m = k;
        ans = 0;
        dfs(0,0,1);
        printf("%I64d
",ans);
    }
    return 0;
}