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。
1 #include<cstdio> 2 #include<iostream> 3 using namespace std; 4 5 6 long long a[15]; 7 long long ans; 8 int cnt; 9 int n,m; 10 11 long long gcd(long long b,long long c)//计算最大公约数 12 { 13 return c==0?b:gcd(c,b%c); 14 } 15 16 long long lcm(long long b,long long c)//计算最小公倍数 17 { 18 return b * c/ gcd(b, c); 19 } 20 21 void dfs(int cur,int num,long long Lcm) 22 //深搜，搜出每一种数学组合的可能，因为m<=10所以不会爆 23 { 24 Lcm=lcm(Lcm,a[cur]); 25 if(num%2==0) 26 ans-=(n-1)/Lcm; 27 else 28 ans+=(n-1)/Lcm; 29 for(int j=cur+1;j<cnt;j++)//这个j只能放在里面定义！！ 30 dfs(j,num+1,Lcm); 31 } 32 //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 33 34 int main() 35 { 36 while(~scanf("%d%d",&n,&m)) 37 { 38 cnt=0; 39 int x; 40 while(m--) 41 { 42 scanf("%d",&x); 43 if(x!=0)//除去0的那种情况 44 a[cnt++]=x; 45 } 46 ans=0; 47 for(int i=0;i<cnt;i++) 48 dfs(i,1,1); 49 //容斥，奇加偶减 50 printf("%lld ",ans); 51 } 52 return 0; 53 }