hdoj-1976-How many integers can you find

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

#include<iostream>
#include<cstdio>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
ll n,m,a[12],ans,p;
ll gcd(ll a,ll b)
{
if (!b) return a;
return gcd(b,a%b);
}
void DFS(ll i,ll w,ll k)
{
for (;i<=n;i++)
if (a[i])
{
p=a[i]*w/gcd(a[i],w);
ans+=k*(m/p);
DFS(i+1,p,-k);
}
return;
}
int main()
{
ll i;
while (cin>>m>>n)
{
for (i=1;i<=n;i++) cin>>a[i];
ans=0;
m--;
DFS(1,1,1);
cout<<ans<<endl;
}
return 0;
}

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。

数据范围比较大，暴力去找肯定超时，2^31数组也存不了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。