HDU - 1796 How many integers can you find

最新推荐文章于 2020-07-06 12:17:07 发布

Lngxling

最新推荐文章于 2020-07-06 12:17:07 发布

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本文介绍了一种解决特定整除计数问题的算法，并通过二进制优化减少了计算层数，有效避免了时间限制超时的问题。文章详细阐述了如何使用二进制表示法来确定是否将某个数包含在集合中，以及如何根据所选数字数量的奇偶性决定增加或减少计数。

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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): 10302    Accepted Submission(s): 3078

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

题意：

给定数n和有m个数的集合

问1-n中有多少数能被集合里的某个数整除

思路：

直接进行容斥有10层会T，要进行二进制优化，即用数的二进制来表示数组里取的和没取的数，再根据取数个数的奇偶来判断是要加还是减

注意要在输入数时要把0去掉...

#include <bits/stdc++.h>
using namespace std;
#define ll long long
ll n;
int m;
int num[20];
int main()
{
	while(scanf("%lld%d",&n,&m)!=EOF)
	{
		n--;
		int pl=0;
		for(int i=1;i<=m;i++)
		{
			scanf("%d",&num[pl]);
			if(num[pl])
			pl++;
		}
		ll ans=0;
		for(int i=1;i<(1<<pl);i++)
		{
			int cnt=0;
			ll sum=1;
			for(int j=0;j<pl;j++)
			{
				if(i&(1<<j))
				{
					cnt++;
					sum=(num[j]/__gcd(sum,(ll)num[j])*sum);
				}
			}
			if(cnt&1)
			ans+=n/sum;
			else
			ans-=n/sum;
		}
		printf("%lld\n",ans);
	}
}