HDU - 1796 How many integers can you find

最新推荐文章于 2019-12-23 00:08:09 发布

原创最新推荐文章于 2019-12-23 00:08:09 发布 · 182 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#新知识

水题同时被 2 个专栏收录

17 篇文章

订阅专栏

容斥定理

1 篇文章

订阅专栏

本篇介绍了一道编程题的解决思路，利用容斥原理找出小于给定整数N的所有能被指定集合中任意整数整除的数的数量。通过实例详细解释了如何运用容斥原理进行计算，并提供了完整的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9580 Accepted Submission(s): 2876

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

大致题意就是，给你一个整数N，和一个集合，找出小于N的被集合里的任意一个整数整除的数有多少个

//第一次接触容斥定理，我看了好久的容斥定理到编码的过程，找了别的题看了看大概明白了，，这道是简单型的还要看看别的

//容斥定理讲解http://blog.youkuaiyun.com/TRiddle/article/details/51873991

所以大致代码借大神的理解着写

注意0不要忘

#include<cstdio>
#include<cmath>
#include<algorithm>
#define LL long long
using namespace std;
LL num[25];
LL n,m;
LL gcd(LL a,LL b)
{
	int r;
	while(b!=0)
	{
	  r=b;
	  b=a%b;
	  a=r;
	}
	return a;
}
LL lcm(LL a,LL b)
{
	return a/gcd(a,b)*b;
}
LL solve()
{
	LL ans = 0;
	for (LL i = 1 ; i <(1<<m) ; i++)
	{
		LL ant = 0;
		LL k = 1;
		for (LL j = 0 ; j < m; j++)
		{
			if (i & (1<<j))
			{
				ant++;
				k=lcm(k,num[j]);//因为m集合内的数不一定是质数，所以当集合内数累乘时应求其最小公倍数。
			}
		}
		if (ant & 1)
			ans += n / k;//奇加偶减
		else
			ans -= n / k;
	}
	return ans;
}
int main()
{
	LL i;
	while(~scanf("%lld %lld",&n,&m))
	{
		n=n-1;//不包括n
	  for(i=0;i<m;i++)
	  {
	    scanf("%lld",&num[i]);
	    if(num[i]==0)   //m集合内的数可能有0，将情况排除
	      {
	      	i--;
	      	m--;
		  }
	  }
	  sort(num,num+m);
	  printf ("%lld\n",solve());
    }
	return 0;
}