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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 <algorithm>
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
#define MOD 1000000007
typedef long long ll;
//判断1-n有多少个M个因子任一个的倍数
int n,m,cnt;
ll ans ,a[30];
ll gcd(ll a,ll b)
{
return b==0?a:gcd(b,a%b);
}
void DFS(int cur,ll lcm,int id)
{// 当前第几个因子 这些因子的最小共公倍数 当前总因子
lcm = a[cur]/gcd(a[cur],lcm)*lcm;
if(id&1) //奇加偶减
ans += (n-1)/lcm;
else
ans -= (n-1)/lcm;
for(int i= cur+1;i<cnt;i++)
DFS(i,lcm,id+1);
}
int main()
{
while(scanf("%d%d",&n,&m) != EOF)
{
cnt = 0;
int x;
while(m--)
{
scanf("%d",&x);
if(x!=0)
a[cnt++] = x;
}
ans = 0;
for(int i=0;i<cnt;i++) //遍历每种组合 从1开始
DFS(i,a[i],1);
printf("%lld\n",ans);
}
return 0;
}