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[];
 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==?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%==)
         ans-=(n-)/Lcm;
     else
         ans+=(n-)/Lcm;
     for(int j=cur+;j<cnt;j++)//这个j只能放在里面定义！！
         dfs(j,num+,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=;
         int x;
         while(m--)
         {
             scanf("%d",&x);
             if(x!=)//除去0的那种情况
                 a[cnt++]=x;
         }
         ans=;
         for(int i=;i<cnt;i++)
             dfs(i,,);
         //容斥，奇加偶减
         printf("%lld\n",ans);
     }
     return ;
 }