How many integers can you find(hdu1796)

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6429    Accepted Submission(s): 1847

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

思路：容斥原理；需要注意的是给你的数有可能包含0，只要把0换成比n-1大的数或者去掉就行；

还有求的是<n的，那么这时麻烦的地方就是要判断整除，所以转变下就是求(<=n-1)就行这时不需要判断是否整除。

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
long long  gcd(long long n,long long  m);
int ans[20];
int main(void)
{
        int i,j,k;
        int n,m;
        while(scanf("%d %d",&n,&m)!=EOF)
        {
                int sum=0;
                n=n-1;
                for(i=0; i<m; i++)
                {
                        scanf("%d",&ans[i]);
                }
                for(i=0; i<m; i++)
                {
                        if(ans[i]==0)
                                ans[i]=n+1;
                }
                for(i=1; i<=(1<<m)-1; i++)
                {
                        int cnt=0;
                        long long  an=1;
                        int flag=0;
                        for(j=0; j<m; j++)
                        {
                                if(i&(1<<j))
                                {
                                        cnt++;
                                        long long  cc=gcd(an,(long long)ans[j]);
                                        an=an/cc*ans[j];
                                        if(an>n)
                                        {
                                                flag=1;
                                                break;
                                        }
                                }
                        }
                        if(flag)
                                continue;
                        else
                        {
                                if(cnt%2)
                                        sum+=n/(int)an;
                                else  sum-=n/(int)an;
                        }
                }
                printf("%d
",sum);
        }
        return 0;
}
long long  gcd(long long  n,long long  m)
{
        if(m==0)
                return n;
        else if(n%m==0)
                return m;
        else return gcd(m,n%m);
}