HDU 1796 How many integers can you find (容斥原理)

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2767    Accepted Submission(s): 778

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

 

 

Author

wangye

 

 

Source

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

 

 

Recommend

wangye

 

 

 

 

题目大意：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

 

解题思路：容斥原理地简单应用。先找出1...n内能被集合中任意一个元素整除的个数，再减去能被集合中任意两个整除的个数，即能被它们两只的最小公倍数整除的个数，因为这部分被计算了两次，然后又加上三个时候的个数，然后又减去四个时候的倍数...所以深搜，最后判断下集合元素的个数为奇还是偶，奇加偶减。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>

using namespace std;

int n,m,cnt;
long long ans,a[30];

long long gcd(long long a,long long b){
    return b==0?a:gcd(b,a%b);
}

void DFS(int cur,long long lcm,int id){
    lcm=a[cur]/gcd(a[cur],lcm)*lcm;
    if(id&1)
        ans+=(n-1)/lcm;     //因为这题并不包含n本身，所以用n-1 
    else
        ans-=(n-1)/lcm;
    for(int i=cur+1;i<cnt;i++)
        DFS(i,lcm,id+1);
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d",&n,&m)){
        cnt=0;
        int x;
        while(m--){
            scanf("%d",&x);
            if(x!=0)    //除0
                a[cnt++]=x;
        }
        ans=0;
        for(int i=0;i<cnt;i++)
            DFS(i,a[i],1);
        cout<<ans<<endl;
    }
    return 0;
}