ACM: How many integers can you find-数论专题-容斥原理的简单应用+GCD

How many integers can you find

Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

 

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

/*/
题意：
给出N和M 输入M个数，找出所有M个数的倍数并且，Mi的倍数小于N，输出所有数的总个数。 

如果一个数同时是三个数的倍数
单独记一个数的倍数次数为C（3，1） =3
记两个数的倍数次数为 C（3，2）=3
记三个数的倍数次数为 C（3，3）=1
3-3+1=1，只记一次依次类推

一个数为5个数的倍数
C（5，1）=5
C（5，2）=10
C（5，3）=10
C（5，4）=5
C（5，5）=1
5-10+10-5+1=1

六个数
C（6，1）=6
C（6，2）=15
C（6，3）=20
C（6，4）=15
C（6，5）=6
C（6，6）=1
6-15+20-15+6-1=1

上图：


然后因为数字不超过10个，可以运用枚举子集的思想去做这个题目。
所以用到DFS。
最后有一个地方要注意就是在DFS里面判断积这里，要用GCD，一开始没想到过不了样例。 

AC代码：
/*/

#include"map"
#include"cmath"
#include"string"
#include"cstdio"
#include"vector"
#include"cstring"
#include"iostream"
#include"algorithm"
using namespace std;
typedef long long LL;

LL a[15];
int n,m,cnt;
LL ans,x;

LL gcd(LL a,LL b){
	return b?gcd(b,a%b):a;
}

void DFS(int x,LL axb,int num) {
	axb=a[x]/gcd(a[x],axb)*axb;
	if(num&1) ans+=(n-1)/axb;
	else ans-=(n-1)/axb;
	//	cout<<"now ans is:"<<ans<<endl;  //检查
	for(int i=x+1; i<cnt; i++)
		DFS(i,axb,num+1);
}

int main() {
	while(~scanf("%d%d",&n,&m)) {
		ans=0;
		cnt=0;
		for(int i=0; i<m; i++) {
			scanf("%I64d",&x);
			if(x!=0)a[cnt++]=x;
		}
		for(int i=0; i<cnt; i++){
			DFS(i,a[i],1);       //用DFS去枚举每种选择的情况。 
		}
		printf("%d
",ans);
	}
	return 0;
}