k很小的时候,能满足的n已经很大了,对k进行遍历,如果(n%i) == i-1,直到最后,那么它就成立
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
int main()
{
__int64 n,k;
cin>>n>>k;
int flag = 1;
for(int i = 1 ; i <= k ;i++)
{
if(n%i == i-1)
continue;
else
{
flag = 0;
break;
}
}
if(flag)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
return 0;
}