C. Cave Painting
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Imp is watching a documentary about cave painting.
Some numbers, carved in chaotic order, immediately attracted his attention. Imp rapidly proposed a guess that they are the remainders of division of a number n by all integers i from 1 to k. Unfortunately, there are too many integers to analyze for Imp.
Imp wants you to check whether all these remainders are distinct. Formally, he wants to check, if all 
, 1 ≤ i ≤ k, are distinct, i. e. there is no such pair (i, j) that:
- 1 ≤ i < j ≤ k,
, where 
is the remainder of division x by y.
Input
The only line contains two integers n, k (1 ≤ n, k ≤ 1018).
Output
Print "Yes", if all the remainders are distinct, and "No" otherwise.
You can print each letter in arbitrary case (lower or upper).
Examples
input
Copy
4 4
output
Copy
No
input
Copy
5 3
output
Copy
Yes
Note
In the first sample remainders modulo 1 and 4 coincide.
题意：问1~k的范围内是否n对其取余的数都不同；
思路：虽然范围很大，但是题目的要求感觉暴力前多少个数就可以判断出来了，因为n对1~k中的数取余的余数要都不同的的话只能是他的值减一，所以范围应该不会很大，我暴力了前1e5个数；
下面附上我的代码：
#include<cstdio>
#include<cmath>
#include<cstring>
#include<map>
#include<set>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<string>
#define LL long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define lson o<<1
#define rson o<<1|1
using namespace std;
int main()
{
LL n, k;
scanf("%lld %lld", &n, &k);
int flag = 0;
for(LL i = 1; i <= 100000 && i <= k; i++)
{
if(n % i != i - 1)
{
puts("No");
flag = 1;
break;
}
}
if(!flag)
puts("Yes");
return 0;
}