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): 6434    Accepted Submission(s): 1849

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这道题目是给定了因子,而且还不是互质的,所以在容斥原理上,就不能简单想乘,要求最小公倍数LCM另外有可能因子为0#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <math.h>

using namespace std;
typedef long long int LL;
LL gcd(LL a,LL b)
{
	return b?gcd(b,a%b):a;
}
LL ans,a[15],n,m;
void dfs(int id,int flag,int lcm)
{
	lcm=a[id]/gcd(a[id],lcm)*lcm;
	if(flag)
		ans+=n/lcm;
	else
		ans-=n/lcm;
	for(int i=id+1;i<m;i++)
		dfs(i,!flag,lcm);
}
int main()
{
	while(cin>>n>>m)
	{
        n--;
		for(int i=0;i<m;i++)
		{
            scanf("%d",&a[i]);
			if(!a[i]) i--,m--;
		}
		ans=0;
		for(int i=0;i<m;i++)
			dfs(i,1,a[i]);
		printf("%lld\n",ans);
	}
	return 0;
}

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