Luo Gu solution to a problem P5535 XR-3] [[gossip]

I double Cheese Li was spicy cabbage!

P5535 [XR-3] grapevine (This question is a flood problem

In casual working of the conscience to remind us:

You may need to use Theorem - Bertrand - cut Theorem Chebyshev.

So what is Bertrand - Chebyshev Theorem?

I do not know, butOmniscient degree of your mother know on the line：

If the integer n> 3, then there is at least one prime number p, comply with n <p <2n - 2. Another argument is slightly weak: for all integer n is greater than 1, there is a prime number p, comply with n <p <2n.

So what's the use of this theorem?

Because it is a number selected from n k, k + 1 must therefore be greater than 1

It is not just and Bertrand - cut coincides conditions than Chebyshev theorem?

Yes, we think about the two cases:

1. If k + 1 is a prime number, then in two ways:

I.1 ~ n multiples do not contain k + 1, then obviously the first day it

II.1 ~ n containing multiple k + 1, then take a few days?

Two days later, Why? Since the first day, all of the multiple number k + 1 is removed all know, so gcd ((2 * (k + 1)), (2 * (k + 1)) + 1) = 1 is constant. So just two days.

2. If k + 1 is not a prime number:

It is two days, to tell people have k + 1 is prime number, then Bertrand -> 3, in line with the presence of p n <p <2 * n n-Chebyshev Theorem;

Konjac may speak not very clear, bear with me, I could understand how the sensibility.

#include<bits/stdc++.h>
using namespace std;

#define int long long //一定要开long long，不开long long 见祖先

int n, k;

bool prime(int x)//很清晰的质数筛 
{
if(x == 1)//特判一下1 
return 0;
if(x == 2)//特判一下2 
return 1;
for(int i = 2; i <= sqrt(x); ++ i)//就是枚举每个数，看看它是否是它的约数 
if(x % i == 0)//如果不是的话 
return 0;//直接return false 
return 1;
}

signed main()
{
scanf("%lld%lld", &n, &k);
if(prime(k + 1) && 2 * k >= n)//其实，也很好理解，想一想，如果一个数是质数那么是不是除去它的倍数的数都与它互质？所以prime(k + 1)是判断它是不是质数，后面就是找有没有它的倍数（因为是连续的所以只有比一下大小即可 
printf("1");
else//除去1的答案就是二了（会有解释 
printf("2");
return 0;
}

PS：Please re-read copy