HDU 1796 How many integers can you find(容斥原理)

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8059    Accepted Submission(s): 2403

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

 

Author

wangye

 

Source

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

 

Recommend

wangye

 

题意:给你一个n和一个m个数的集合，求1到n中满足能被集合中一个数整除的数有几个

思路:根据容斥定理，满足条件的数为被集合中一个数整除的数的数量和-能被集合中任意两个数整除的数的数量+三个......

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;typedef long long ll;int a[20];int n,m;void solve(){    int res=0;    for(int i=1;i<(1<<m);i++)    {        int num=0;        for(int j=i;j!=0;j>>=1) num+=j&1;        int lcm=1;        for(int j=0;j<m;j++)        {            if(i>>j&1)            {                lcm=lcm/__gcd(lcm,a[j])*a[j];                if(lcm>n) break;            }        }        if(num%2==0) res-=(n/lcm);        else res+=(n/lcm);    }    printf("%d\n",res);}int main(){    while(~scanf("%d%d",&n,&m))    {        n--;        int flag=1;        for(int i=0;i<m;i++)        {            scanf("%d",&a[i]);            if(a[i]==0)            {                i--;m--;            }        }        solve();    }    return 0;}