B. Vile Grasshoppers
http://codeforces.com/problemset/problem/937/B
The weather is fine today and hence it's high time to climb the nearby pine and enjoy the landscape.
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
.
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.
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.
The only line contains two integers p and y (2 ≤ p ≤ y ≤ 109).
Output the number of the highest suitable branch. If there are none, print -1 instead.
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.
It immediately follows that there are no valid branches in second sample case.
水题
// 去吧!皮卡丘! 把AC带回来!
// へ /|
// /\7 ∠_/
// / │ / /
// │ Z _,< / /`ヽ
// │ ヽ / 〉
// Y ` / /
// イ● 、 ● ⊂⊃〈 /
// () へ | \〈
// >ー 、_ ィ │ //
// / へ / ノ<| \\
// ヽ_ノ (_/ │//
// 7 |/
// >―r ̄ ̄`ー―_
//**************************************
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 2147483647
const ll INF = 0x3f3f3f3f3f3f3f3fll;
#define ri register int
template <class T> inline T min(T a, T b, T c) { return min(min(a, b), c); }
template <class T> inline T max(T a, T b, T c) { return max(max(a, b), c); }
template <class T> inline T min(T a, T b, T c, T d) {
return min(min(a, b), min(c, d));
}
template <class T> inline T max(T a, T b, T c, T d) {
return max(max(a, b), max(c, d));
}
#define scanf1(x) scanf("%d", &x)
#define scanf2(x, y) scanf("%d%d", &x, &y)
#define scanf3(x, y, z) scanf("%d%d%d", &x, &y, &z)
#define scanf4(x, y, z, X) scanf("%d%d%d%d", &x, &y, &z, &X)
#define pi acos(-1)
#define me(x, y) memset(x, y, sizeof(x));
#define For(i, a, b) for (ll i = a; i <= b; i++)
#define FFor(i, a, b) for (ll i = a; i >= b; i--)
#define bug printf("***********\n");
#define mp make_pair
#define pb push_back
const int maxn = 3e5 + ;
const int maxx = 1e6 + ;
// name*******************************
ll p, y;
// function****************************** //***************************************
int main() {
cin >> p >> y;
bool flag = true;
FFor(i, y, p + ) {
flag = true;
for (ll j = ; j <= p && j * j <= i; j++) {
if (i % j == ) {
flag = false;
break;
}
}
if (flag) {
cout << i;
return ;
}
}
cout<<-; return ;
}
B. Vile Grasshoppers的更多相关文章
- Codeforces Round #467 (Div. 2) B. Vile Grasshoppers
2018-03-03 http://codeforces.com/problemset/problem/937/B B. Vile Grasshoppers time limit per test 1 ...
- Codeforces 937.B Vile Grasshoppers
B. Vile Grasshoppers time limit per test 1 second memory limit per test 256 megabytes input standard ...
- Codeforces Round #467 (Div. 2) B. Vile Grasshoppers[求去掉2-y中所有2-p的数的倍数后剩下的最大值]
B. Vile Grasshoppers time limit per test 1 second memory limit per test 256 megabytes input standard ...
- CodeForces937B:Vile Grasshoppers(素数性质)
The weather is fine today and hence it's high time to climb the nearby pine and enjoy the landscape. ...
- A - Vile Grasshoppers
Problem description The weather is fine today and hence it's high time to climb the nearby pine and ...
- CF937B Vile Grasshoppers
Vile Grasshoppers time limit per test 1 second memory limit per test 256 megabytes input standard in ...
- Codeforces Round #467 (div.2)
Codeforces Round #467 (div.2) 我才不会打这种比赛呢 (其实本来打算打的) 谁叫它推迟到了\(00:05\) 我爱睡觉 题解 A. Olympiad 翻译 给你若干人的成绩 ...
- codeforce round #467(div.2)
A. Olympiad 给出n个数,让你找出有几个非零并且不重复的数 所以用stl的set //#define debug #include<stdio.h> #include<ma ...
- 【codeforces】【比赛题解】#937 CF Round #467 (Div. 2)
没有参加,但是之后几天打了哦,第三场AK的CF比赛. CF大扫荡计划正在稳步进行. [A]Olympiad 题意: 给\(n\)个人颁奖,要满足: 至少有一个人拿奖. 如果得分为\(x\)的有奖,那么 ...
随机推荐
- druapl7:"Notice: A non well formed numeric value encountered 在 _hierarchical_select_hierarchy_generate() "
这个是很诡异的一个Notice错误提醒,因为我在Drupal7.54+PHP7.0.1的环境下,并没有报这个错.但是我再Drupal7.59+PHP7.1.7的环境下就报错了.很奇怪,按照报错信息bi ...
- call aplly笔记
<script> /*1.每个函数都包含两个非继承而来的方法:apply()和call(). 2.他们的用途相同,都是在特定的作用域中调用函数. 3.接收参数方面不同,apply()接收两 ...
- 原生爬虫小Demo
import re from urllib import request class Spider(): url = 'https://www.panda.tv/cate/lol' #[\s\S]匹配 ...
- Android8.0适配那点事(一)
最近有小伙伴说,7.0适配整了一波,现在又要来适配8.0,真是一波未平一波又起 但是作为开发者来说,学无止境,不跟上时代的步伐,肯定会被时代所淘汰... 话说Android P已经在路上了,你准备好了 ...
- java基础(七) java四种访问权限
引言 Java中的访问权限理解起来不难,但完全掌握却不容易,特别是4种访问权限并不是任何时候都可以使用.下面整理一下,在什么情况下,有哪些访问权限可以允许选择. 一.访问权限简介 访问权限控制: ...
- mysql group replication观点及实践
一:个人看法 Mysql Group Replication 随着5.7发布3年了.作为技术爱好者.mgr 是继 oracle database rac 之后. 又一个“真正” 的群集,怎么做到“ ...
- cef开启摄像头和录音
参考资料:https://github.com/cztomczak/phpdesktop/wiki/Chrome-settings#command_line_switches CefSharp中文帮助 ...
- 动态展开tableView的cell[2]
动态展开tableView的cell[2] http://code4app.com/ios/%E5%8A%A8%E6%80%81%E6%B7%BB%E5%8A%A0cell/53845f8a933bf ...
- 前端 网络三剑客之html 02
html 四.表单标签 form标签: input系列:内敛标签 1.明文: 姓名:<input type="text" name="user" plac ...
- 资料整理,SQL Server ,面试前复习笔记
T-SQL 要掌握的知识点分类 SQL 面向数据库执行查询 SQL 从数据库取回数据 SQL 在数据库中插入新的记录 SQL 更新数据库中的数据 SQL 从数据库删除记录 SQL 创建新数据库 SQL ...