Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) - A
题目链接:http://codeforces.com/contest/831/problem/A
题意:给定一个序列,问你这个序列是否是单峰的。 定义单峰的序列为: (序列值的变化趋势)开始是递增的,然后是平缓的,最后是递减的。
对于开始(递增)和结尾(递减)的那部分可以不出现
思路:按照题目判即可。
import java.io.*;
import java.util.*; public class Main {
public static final int MAXN=100+24;
public static int n;
public static int[] val=new int[MAXN];
public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
PrintWriter out=new PrintWriter(System.out);
n=cin.nextInt();
for(int i=1;i<=n;i++){
val[i]=cin.nextInt();
}
int up=n+1,down=0;
for(int i=1;i<n;i++){
if(val[i] >= val[i+1]){
up=i; break;
}
}
for(int i=n;i>1;i--){
if(val[i] >= val[i-1]){
down=i; break;
}
}
boolean flag=true;
for(int i=up;i<down;i++){
if(val[i]!=val[i+1]){
flag=false;
break;
}
}
out.println(flag?"YES\n":"NO\n");
out.flush(); out.close(); cin.close();
}
}
Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) - A的更多相关文章
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals)
http://codeforces.com/contest/831 A. Unimodal Array time limit per test 1 second memory limit per te ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem C (Codeforces 831C) - 暴力 - 二分法
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals)A,B,C
A:链接:http://codeforces.com/contest/831/problem/A 解题思路: 从前往后分别统计递增,相等,递减序列的长度,如果最后长度和原序列长度相等那么就输出yes: ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem F (Codeforces 831F) - 数论 - 暴力
题目传送门 传送门I 传送门II 传送门III 题目大意 求一个满足$d\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d} \right \rceil - \sum ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem D (Codeforces 831D) - 贪心 - 二分答案 - 动态规划
There are n people and k keys on a straight line. Every person wants to get to the office which is l ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem E (Codeforces 831E) - 线段树 - 树状数组
Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this int ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem A - B
Array of integers is unimodal, if: it is strictly increasing in the beginning; after that it is cons ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) A 水 B stl C stl D 暴力 E 树状数组
A. Unimodal Array time limit per test 1 second memory limit per test 256 megabytes input standard in ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) - D
题目链接:http://codeforces.com/contest/831/problem/D 题意:在一个一维坐标里,有n个人,k把钥匙(钥匙出现的位置不会重复并且对应位置只有一把钥匙),和一个终 ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) - C
题目链接:http://codeforces.com/contest/831/problem/C 题意:给定k个评委,n个中间结果. 假设参赛者初始分数为x,按顺序累加这k个评委的给分后得到k个结果, ...
随机推荐
- 170826-关于spring的知识点及练习
1.Spring作用: 1.生态体系庞大,全能型选手![springmvc是其一个子模块,jdbcTemplate能直接操作数据库!] 2.将其他组件粘合在一起 3.IOC容器和AOP[Aspect ...
- [洛谷P3938]:斐波那契(fibonacci)(数学)
题目传送门 题目描述 小$C$养了一些很可爱的兔子.有一天,小$C$突然发现兔子们都是严格按照伟大的数学家斐波那契提出的模型来进行繁衍:一对兔子从出生后第二个月起,每个月刚开始的时候都会产下一对小兔子 ...
- php简单随机实现发红包程序
前言: 使用PHP发红包,当我们输入红包数量和总金额后,PHP会根据这两个值进行随机分配每个金额,保证每个人都能领取到一个红包,每个红包金额不等,就是要求红包金额要有差异,所有红包金额总额应该等于总金 ...
- Linux下开启FTP服务
一.配置步骤 1.安装vsftp 使用yum命令安装vsftp #yum install vsftpd -y 2.添加ftp帐号和目录 先确定nologin的位置,通常在/usr/sbin/nolog ...
- canvas万花筒案例
<!DOCTYPE html><html><head> <meta charset="UTF-8"> <title>Ti ...
- 用Vue来实现音乐播放器(五):路由配置+顶部导航栏组件开发
路由配置 在router文件夹下的index.js中配置路由 import Vue from 'vue' import Router from 'vue-router'//配置路由前先引入组件impo ...
- 测开之路八十六:python操作sqlite
创建sqlite数据库,并创建表和数据 python自带sqlite3库可以创建数据库文件 导入库:import sqlite3 创建游标,指定数据库名字:con = sqlite3.connect( ...
- postman+newman+jenkins接口自动化
postman用来做接口测试非常方便,接口较多时,则可以实现接口自动化 目录 1.环境准备 2.本机调试脚本 3.集成jenkins 1.环境准备 1.1安装nodejs6.0+ 安装nodejs6. ...
- Git 实践
最近也学习了Git的相关知识,现通过一个实例来记录Git使用流程,也方便日后使用. git的基础学习: https://www.yiibai.com/git/git-quick-start.html ...
- ubuntu环境下重启mysql服务报错“No directory, logging in with HOME=-”
前提:使用系统的环境 3.13.0-24-generic mysql的版本:5.6.33 错误描述: 首先用mysqld_safe启动报错如下: root@zabbix-forFunction:~# ...