The weather is fine today and hence it's high time to climb the nearby pine and enjoy the landscape.c++
The pine's trunk includes several branches, located one above another and numbered from 2 to y. Some of them (more precise, from 2 to p) are occupied by tiny vile grasshoppers which you're at war with. These grasshoppers are known for their awesome jumping skills: the grasshopper at branch x can jump to branches 
.ssh
Keeping this in mind, you wisely decided to choose such a branch that none of the grasshoppers could interrupt you. At the same time you wanna settle as high as possible since the view from up there is simply breathtaking.ide
In other words, your goal is to find the highest branch that cannot be reached by any of the grasshoppers or report that it's impossible.ui
The only line contains two integers p and y (2 ≤ p ≤ y ≤ 109).this
Output the number of the highest suitable branch. If there are none, print -1 instead.spa
3 6
5
3 4
-1
In the first sample case grasshopper from branch 2 reaches branches 2, 4 and 6 while branch 3 is initially settled by another grasshopper. Therefore the answer is 5.code
It immediately follows that there are no valid branches in second sample case.blog
[分析]: 给你两个数p，y，求去掉2-y中全部2-p的数的倍数后剩下的最大值，若是没有剩下数输出-1.ci
[代码]:和筛法一个意思,注意最大值能够逆向枚举.只要算到sqrt（10^9）（i<=p这个条件的加入是为了防止x可能不是素数，可是x不含有2~p之间的任何一个因数）input
[代码]:
#include<bits/stdc++.h> using namespace std; int p, y; bool ck(int n) { for(int i = 2; i * i <= n && i <= p; i++) { if(n % i == 0) { return false; } } return true; } int main() { cin >> p >> y; for(int i = y; i > p; i--) { if(ck(i)) { printf("%d\n", i); return 0; } } puts("-1"); return 0; }