【HDU】1796 - How many integers can you find（容斥原理，GCD）

点击打开题目

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6760 Accepted Submission(s): 1961

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

Source

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

有几点需要注意：

①m里的值可能是0

②m的值可能不互质，这时候取数的时候不能直接相乘，要求LCM( )

③m里的数可能大于等于n

④n要减一

代码如下：

#include <cstdio>
#include <cstring>
int GCD(int a,int b)
{
	if (a % b == 0)
		return b;
	return GCD(b,a%b);
}
int LCM(int a,int b)
{
	return a/GCD(a,b)*b;
}
int main()
{
	int n;
	int m;
	int num[30];
	while (~scanf ("%d %d",&n,&m))
	{
		for (int i = 0 ; i < m ; i++)
		{
			scanf ("%d",&num[i]);
			if (num[i] < 1 || num[i] >= n)		//大于等于n的也要删掉 
			{
				m--;
				i--;
			}
		}
		//用容斥原理直接求
		n--;
		int ans = 0;
		for (int i = 1 ; i < (1 << m) ; i++)
		{
			int v = 1;
			int k = 0;		//计数
			for (int j = 0 ; j < m ; j++)
			{
				if ((1 << j) & i)
				{
					v = LCM(v,num[j]);
					k++;
				}
			}
			if (k & 1)
				ans += n / v;
			else
				ans -= n / v;
		}
		printf ("%d\n",ans);
	}
	return 0;
}