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

2024-10-29 14:45:02   207人阅读

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


题目大意：给出一个n和一个集合，m个数，问在1到n内 能被m中的数正除的有多少？

可以理解为在1到n内，是m中的数的倍数的数有多少？  

在容斥原理中，用二进制数表示第i个数取没取，如果取了奇数个数，加上。 取了偶数个数，减去

#include <cstdio>
#include <cstring>
#include <algorithm>
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<<j & i )
                {
                    k = a[j] * k / ( gcd(k,a[j]) ) ;
                    num++ ;
                }
            }
            if( num % 2 )
            {
                ans += n/k ;
                if( n % k == 0 )
                    ans-- ;
            }
            else
            {
                ans -= n/k ;
                if(n%k == 0)
                    ans++ ;
            }
        }
        printf("%lld\n", ans) ;
    }
    return 0;
}


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


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