hdoj How many integers can you find 1796 （容斥原理&&DFS）

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 5994    Accepted Submission(s): 1719

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
  
  
    
  
  
   思路：
  
  
   先找出從1到n之間能被給定集合内的元素整除的數的個數，再減去能被兩個元素同時整除的數的個數，再加上能同時被三個數整除的數的個數，再減去.........;通過這樣可以發現，“奇加偶減”這一規律。運用dfs完成。
  
  
          
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[30];
int kk;
int ans;
int n,m;
long long gcd(long long a,long long b)
{
	return b==0?a:gcd(b,a%b);
}
void dfs(int k,int x,int d)
{
	x=a[k]/gcd(a[k],x)*x;
	if(d&1)
		ans+=(n-1)/x;
	else
		ans-=(n-1)/x;
	for(int i=k+1;i<kk;i++)
		dfs(i,x,d+1);
}
int main()
{
	int i,j,x;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		kk=0;
		while(m--)
		{
			scanf("%d",&x);
			if(x!=0)
				a[kk++]=x;
		}
		ans=0;
		for(i=0;i<kk;i++)
			dfs(i,a[i],1);
		printf("%d\n",ans);
	}
	return 0;
}