C. Cave Painting
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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
inputCopy
4 4
output
No
inputCopy
5 3
output
Yes
Note
In the first sample remainders modulo 1 and 4 coincide.
题意：给定一个n，k，如果n mod i（1<=i<=k）是不同的，输出Yes,否则输出No。
思路：看到数据范围下了一跳，想了很久数论，但好像都行不通。最后暴力了一发，AC？
首先n%1=0，那么n%2取值只有0，1，如果是0那么No，所以n%2取值只能是1，递推下去……找到规律，若要Yes，n%i只能为i-1。仔细想想，能够Yes的n和k应该都不会很大。暴力即可。
代码：
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define met(s,k) memset(s,k,sizeof s)
#define scan(a) scanf("%d",&a)
#define scanl(a) scanf("%lld",&a)
#define scann(a,b) scanf("%d%d",&a,&b)
#define scannl(a,b) scanf("%lld%lld",&a,&b)
#define scannn(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define prin(a) printf("%d\n",a)
#define prinl(a) printf("%lld\n",a)
using namespace std;
typedef long long ll;
const int maxn=505*505;
int main()
{
ll n,k;
int flag=0;
scannl(n,k);
if(k>=n&&n!=1)
{
printf("No\n");
return 0;
}
for(int i=2;i<=k;i++)
{
if(n%i!=i-1)
{
printf("No\n");
flag=1;
break;
}
}
if(!flag)printf("Yes\n");
return 0;
}