hdu1796—How many integers can you find（简单容斥）

题目链接：传送门

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8740    Accepted Submission(s): 2594

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 22 3

 

Sample Output

7

 

第一道容斥纪念

#include <iostream>#include <cstring>#include <cstdio>using namespace std;typedef long long ll;const int MAXN = 515;const int INF = 0x3f3f3f3f;int st[30];ll gcd(ll a,ll b){    return b==0?a:gcd(b,a%b);}ll lcm(ll a,ll b){    return a/gcd(a,b)*b;}ll Cal( int n , int m ){    ll sum = 0;    for( int i = 1 ; i < (1<<n) ; ++i ){        ll mult=1,bits=0;        for( int j = 0 ; j < n ; ++j ){            if(i&(1<<j)){                bits++;                mult = lcm(mult,st[j]);            }        }        ll cur = (ll)m/mult;        if( bits%2 == 1 ) sum += cur;        else sum -= cur;    }    return sum;}int main(){    int n,m;    while( ~scanf("%d%d",&n,&m) ){        int a,cnt = 0;        for( int i = 0 ; i < m ; ++i ){            scanf("%d",&a);            if( a == 0 ) continue;            st[cnt++] = a;        }        printf("%lld\n",Cal(cnt,n-1));    }    return 0;}