How many integers can you find（容斥+dfs)

最新推荐文章于 2019-08-15 21:04:02 发布

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How many integers can you find

Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5653 Accepted Submission(s): 1625

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

Author

wangye

Source

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

Recommend

wangye

<span style="font-size:18px;">#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
int a[12],cnt,n,m;
LL ans;
LL gcd(LL a,LL b) { return (b==0?a:gcd(b,a%b)); }
void dfs(int i,LL lcm,int id)
{
    lcm=(a[i]*lcm)/gcd(a[i],lcm);
    if(id&1)
        ans +=(n-1)/lcm;
    else
        ans -=(n-1)/lcm;
    for(int j=i+1;j<cnt;j++)
        dfs(j,lcm,id+1);
}
int main()
{
    int x;
    while(scanf("%d %d",&n,&m) != EOF)
    {
        cnt=0,ans=0;
        for(int i=0;i<m;i++)
        {
            scanf("%d",&x);
            if(x!=0)
                a[cnt++]=x;
        }
        for(int i=0;i<cnt;i++)
            dfs(i,a[i],1);
        printf("%lld\n",ans);
    }
    return 0;
}
</span>

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。