CF div2 C. Cave Painting

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：

给你2个数n,k 问是否1到k modn的结果都不一样

思路：

我们可以发现1取余任何数都为0,2取余任何数都只有结果0和1，又因为我们要每个数取余n结果都不一样，故每个数取余n的结果只能是这个数-1.

CODE:

#include<stdio.h> #include<iostream> #include<vector> #include<queue> #include<algorithm> using namespace std; typedef long long LL; int main() { LL n,k,flag=1,t=1000000; cin>>n>>k; t=min(t,k); for(LL i=1;i<=t;i++) { if(n%i!=i-1) { flag=0; break; } } if(flag) cout<<"Yes"<<endl; else cout<<"No"<<endl; return 0; }