HDU 1796 How many integers can you find【容斥原理】

最新推荐文章于 2019-08-11 12:00:10 发布

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本文探讨了一个经典的整除问题：给定一个数N和一个整数集合，找出小于N且能被集合中任意整数整除的所有整数。通过使用容斥原理，文章提供了一种有效的算法解决方案，并附带了完整的C++代码实现。

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How many integers can you find

Time Limit : 12000/5000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 96 Accepted Submission(s) : 31

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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）

容斥原理：

奇数为+，偶数为-

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
#define ll long long
#define ms(a,b)  memset(a,b,sizeof(a))
const int M=1e5+10;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
int i,j,k,n,m;
int cnt;
long long a[M];
int ans;

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

void dfs(int cur,ll lcm,int id)
{
    lcm=a[cur]/gcd(a[cur],lcm)*lcm;
    if(id&1)
        ans+=(n-1)/lcm;
    else ans-=(n-1)/lcm;
    for(int i=cur+1;i<cnt;i++)
        dfs(i,lcm,id+1);
}
int main()
{
    while(~scanf("%d%d",&n,&m)){
        cnt=0;
        for(int i=0;i<m;i++){
            int x;
            scanf("%d",&x);
            if(x!=0)a[cnt++]=x;
        }
        ans=0;
        for(int i=0;i<cnt;i++){
            dfs(i,a[i],1);
        }
        printf("%d\n",ans);
    }
    return 0;
}