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

 

Sample Output

 

Author

wangye

 

Source

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

 

题目大意：很简单的题目，直接看意思就懂哈！

      解题思路：容斥定理，加奇减偶，开始忘记求lcm了，囧！！而且开始还特判0的情况，题目中说的必须是除以，所以0不是一个解。。。开始竟然以为需要是因子就可以了。想通了之后直接先筛选一次，把0都筛选出去。

      题目地址：How many integers can you find

AC代码：

#include<iostream>

#include<cstring>

#include<string>

#include<cmath>

#include<cstdio>

using namespace std;

__int64 sum;

int n,m;

int a[25];

int b[25];

int visi[25];

__int64 gcd(__int64 m,__int64 n)

{

    __int64 tmp;

    while(n)

    {

        tmp=m%n;

        m=n;

        n=tmp;

    }

    return m;

}

__int64 lcm(__int64 m,__int64 n)

{

    return m/gcd(m,n)*n;

}

void cal()

{

    int flag=0,i;

    __int64 t=1;

    __int64 ans;

    for(i=0;i<m;i++)

    {

        if(visi[i])

        {

            flag++;   //记录用了多少个数

            t=lcm(t,b[i]);

        }

    }

    ans=n/t;

    if(n%t==0) ans--;

    if(flag&1) sum+=ans;   //加奇减偶

    else sum-=ans;

}

int main()

{

    int i,j,p;

    while(~scanf("%d%d",&n,&m))

    {

        sum=0;

        for(i=0;i<m;i++)

            scanf("%d",&a[i]);

        int tt=0;  //

        for(i=0;i<m;i++)

        {

            if(a[i])  //去掉0

                b[tt++]=a[i];

        }

        m=tt;

        p=1<<m;   //p表示选取多少个数，组合数的状态

        for(i=1;i<p;i++)

        {

            int tmp=i;

            for(j=0;j<m;j++)

            {

                visi[j]=tmp&1;

                tmp>>=1;

            }

            cal();

        }

        printf("%I64d\n",sum);

    }

    return 0;

}

/*

12 2

2 3

12 3

2 3 0

12 4

2 3 2 0

*/

//968MS