HDU 1796 How many integers can you find

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5605    Accepted Submission(s): 1602

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 2 2 3

 

Sample Output

7

 

Author

wangye

 

容斥，是上一题的简化版.注意M个数 可以有0，有0 的时候要把0去掉

/* ***********************************************
Author        :pk29
Created Time  :2015/8/20 10:01:50
File Name     :4.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

bool cmp(int a,int b){
    return a>b;
}
ll n,m;
ll arr[20];
ll gcd(ll a,ll b){
    return a==0?b:gcd(b%a,a);
}
ll lcm(ll a,ll b){
    ll c=gcd(a,b);
    return a*b/c;
}
ll solve(){
    ll sum=0;
    n--;
    for(int i=1;i<(1<<m);i++){
        int bits=0;
        ll mul=1;
        for(int j=0;j<m;j++){
            if(i&(1<<j)){
                bits++;
                mul=lcm(arr[j],mul);
            }
        }
        if(bits&1)sum+=n/mul;
        else sum-=n/mul;
    }
    return sum;
}
int main()
{
    #ifndef ONLINE_JUDGE
    //freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);

    while(cin>>n>>m){
        int cnt=0,x;
        for(int i=0;i<m;i++){
            cin>>x;
            if(x!=0)arr[cnt++]=x;
        }
        m=cnt--;
        cout<<solve()<<endl;
    }
    return 0;
}