Hdu--1796（数论，容斥原理）

2014-09-05 01:01:28

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4202    Accepted Submission(s): 1196

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

 

思路：经典的容斥原理，枚举+容斥，（任1个数的公倍数） - （任2个数的公倍数） + （任3个数的公倍数）。。。。。

/*************************************************************************
    > File Name: hdu1796.cpp
    > Author: Nature
    > Mail: 564374850@qq.com 
    > Created Time: Thu 04 Sep 2014 09:53:53 PM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
int Gcd(int a,int b) { return b == 0 ? a : Gcd(b,a % b); }
int Lcm(int a,int b) { return a / Gcd(a,b) * b; }

int n,m;
int v[15];
int ans;

void Dfs(int p,int cnt,int l,int tar){
    if(p > m){
        //printf("l : %d,tar : %d\n",l,tar);
        if(cnt == tar){
            int tmp = (n - 1) / l;
            ans += (tar & 1) ? tmp : -tmp;
        }
        return;
    }
    if(m - p + 1 + cnt < tar)
        return;
    Dfs(p + 1,cnt,l,tar);
    int tl = Lcm(l,v[p]);
    if(tl < n && cnt < tar)
        Dfs(p + 1,cnt + 1,tl,tar);
}

int main(){
    while(scanf("%d %d",&n,&m) != EOF){
        for(int i = 1; i <= m; ++i){
            scanf("%d",&v[i]);
            if(!v[i]) --i,--m;
        }
        ans = 0;
        for(int i = 1; i <= m; ++i)
            Dfs(1,0,1,i);
        printf("%d\n",ans);
    }
    return 0;
}