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hdu1796--How many integers can you find（容斥原理）

編輯：C++入門知識

hdu1796--How many integers can you find（容斥原理）

How many integers can you find Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Appoint description:

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 Output

For each case, output the number.

Sample Input

 12 2
2 3

Sample Output

題目大意：給出一個n和一個集合，m個數，問在1到n內能被m中的數正除的有多少？

可以理解為在1到n內，是m中的數的倍數的數有多少？

在容斥原理中，用二進制數表示第i個數取沒取，如果取了奇數個數，加上。取了偶數個數，減去

#include 
#include 
#include 
using namespace std ;
#define LL long long
LL a[12] ;
LL n , m , k , i , j , cnt , num , ans ;
LL gcd(LL x,LL y)
{
    return x == 0 ? y : gcd(y%x,x) ;
}
int main()
{
    while(scanf("%lld %lld", &n, &m) != EOF)
    {
        ans = 0 ;
        for(i = 0 ; i < m ; i++)
        {
            scanf("%lld", &a[i]) ;
            if( a[i] == 0 )
            {
                i-- ;
                m-- ;
            }
        }
        cnt = 1 << m ;
        for(i = 1 ; i < cnt ; i++)
        {
            k = 1 ;
            num = 0 ;
            for(j = 0 ; j < m ; j++)
            {
                if( 1<