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

题目传送：http://acm.hdu.edu.cn/diy/contest_showproblem.php?cid=20918&pid=1002

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

启发博客：http://www.cnblogs.com/jackge/archive/2013/04/03/2997169.html

题目大意：给定n和一个大小为m的集合，集合元素为非负整数。为1...n内能被集合里任意一个数整除的数字个数。n<=2^31,m<=10

 

解题思路：容斥原理地简单应用。先找出1...n内能被集合中任意一个元素整除的个数，再减去能被集合中任意两个整除的个数，即能被它们两只的最小公倍数整除的个数，因为这部分被计算了两次，然后又加上三个时候的个数，然后又减去四个时候的倍数...所以深搜，最后判断下集合元素的个数为奇还是偶，奇加偶减。

#include<cstdio>
#include<iostream>
using namespace std;


long long a[15];
long long ans;
int cnt;
int n,m;

long long gcd(long long b,long long c)//计算最大公约数
{
    return c==0?b:gcd(c,b%c);
}

long long lcm(long long b,long long c)//计算最小公倍数
{
    return b * c/ gcd(b, c);  
}

void dfs(int cur,int num,long long Lcm)
//深搜，搜出每一种数学组合的可能，因为m<=10所以不会爆
{
    Lcm=lcm(Lcm,a[cur]);
    if(num%2==0)
        ans-=(n-1)/Lcm;
    else
        ans+=(n-1)/Lcm;
    for(int j=cur+1;j<cnt;j++)//这个j只能放在里面定义！！
        dfs(j,num+1,Lcm);
}
//cur指当前数字在数组中的位置,num指目前计算公倍数的数字是偶是奇,Lcm指目前计算出的最小公倍数

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        cnt=0;
        int x;
        while(m--)
        {
            scanf("%d",&x);
            if(x!=0)//除去0的那种情况
                a[cnt++]=x;
        }
        ans=0;
        for(int i=0;i<cnt;i++)
            dfs(i,1,1);
        //容斥，奇加偶减
        printf("%lld
",ans);
    }
    return 0;
}