HDU 1796 How many integers can you find

How many integers can you find

Time Limit: 5000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 1796
64-bit integer IO format: %I64d      Java class name: Main

 

  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

Source

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

 

解题：容斥原理，注意有0

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 20;
int a[maxn],n,m;
LL LCM(LL a,LL b){
    return a/__gcd(a,b)*b;
}
int main(){
    while(~scanf("%d%d",&n,&m)){
        int tot = m;
        for(int i = m = 0,tmp; i < tot; ++i){
            scanf("%d",&tmp);
            if(tmp) a[m++] = tmp;
        }
        LL ret = 0;
        for(int i = 1; i < (1<<m); ++i){
            LL lcm = 1;
            int cnt = 0;
            bool flag = false;
            for(int j = 0; j < m; ++j){
                if((i>>j)&1){
                    cnt++;
                    lcm = LCM(lcm,a[j]);
                    if(lcm < 0){
                        flag = true;
                        break;
                    }
                }
            }
            if(flag) continue;
            if(cnt&1) ret += (n - 1)/lcm;
            else ret -= (n - 1)/lcm;
        }
        printf("%I64d
",ret);
    }
    return 0;
}