Codeforces Round #424 A(模拟)
#include<cstdio>
int n,a[];
int main(){
scanf("%d",&n);
for(int i=;i<=n;++i)scanf("%d",a+i);
int L=,R=n;
while(L<n&&a[L]<a[L+])++L;
while(R>&&a[R]<a[R-])--R;
for(int i=L+;i<=R;++i)if(a[L]!=a[i])return puts("NO"),;
return puts("YES"),;
}
#include <iostream>
using namespace std;
int N;
int V[];
int main()
{
cin>>N;
for(int i=;i<=N;i++)cin>>V[i];
int i=;
while(i<N&&V[i]<V[i+])i++;
while(i<N&&V[i]==V[i+])i++;
while(i<N&&V[i]>V[i+])i++;
if(i<N)cout<<"NO";
else cout<<"YES";
return ;
}
#include<iostream>
#include<cstdio>
using namespace std;
int n;
int a[];
int main()
{
scanf("%d",&n);
for(int i=;i<=n;i++)scanf("%d",a+i);
int i=;
while(i+<=n && a[i+]>a[i])i++;
while(i+<=n && a[i+]==a[i])i++;
while(i+<=n && a[i+]<a[i])i++;
if(i<n)printf("NO");else printf("YES");
return ;
}
#include<bits/stdc++.h>
using namespace std; int main()
{
int n;
cin>>n;
int arr[n];
for(int i=;i<n;i++)
cin>>arr[i];
int i=;
while(arr[i+]>arr[i] && i+<n)
i++;
while(arr[i]==arr[i+] && i+<n)
i++;
while(arr[i]>arr[i+] && i+<n)
i++;
if(i==n-)
cout<<"YES";
else cout<<"NO"; }
#include <bits/stdc++.h> using namespace std; int a[];
int n;
void sol(string s) {
cout << s;
exit();
}
int main() {
#ifndef ONLINE_JUDGE
freopen("CF.in", "r", stdin);
#endif
cin >> n;
for (int i = ; i <= n; ++i)cin >> a[i];
int i=;
while(i+<=n&&a[i]<a[i+])++i;
while(i+<=n&&a[i]==a[i+])++i;
while(i+<=n)
{
if(a[i]<=a[i+])sol("NO");
++i;
}sol("YES");
}
Codeforces Round #424 A(模拟)的更多相关文章
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) - B
题目链接:http://codeforces.com/contest/831/problem/B 题意:给第2个26个字母并不重复的字符串(2个字符串对于一个映射),第1个字符串为key集合,第2个字 ...
- 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)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$d\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d} \right \rceil - \sum ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem D (Codeforces 831D) - 贪心 - 二分答案 - 动态规划
There are n people and k keys on a straight line. Every person wants to get to the office which is l ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem E (Codeforces 831E) - 线段树 - 树状数组
Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this int ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem C (Codeforces 831C) - 暴力 - 二分法
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem A - B
Array of integers is unimodal, if: it is strictly increasing in the beginning; after that it is cons ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) E. Cards Sorting 树状数组
E. Cards Sorting time limit per test 1 second memory limit per test 256 megabytes input standard inp ...
随机推荐
- BZOJ4448 SCOI2015情报传递(离线+树链剖分+树状数组)
即滋磁单点修改,询问路径上小于某数的值有多少个.暴力树剖套个主席树(或者直接树上主席树,似乎就1个log了?感觉不一定比两个log快)即可,然而不太优美. 开始觉得可以cdq,然而就变成log^3了. ...
- [洛谷P2044][NOI2012]随机数生成器
题目大意:给你$m,a,c,X_0,n,g$,求$X_{n+1}=(a\cdot X_n+c) \bmod{m}$,最后输出对$g$取模 题解:矩阵快速幂+龟速乘,这里用了$long\;double$ ...
- Flink 任务打包、提交
一.Flink版本 flink-1.6.1-bin-hadoop26-scala_2.11 二.Flink任务打包 笔者将写好的flink计算任务代码发到服务器(ubuntu16.04),在服务器端进 ...
- 线程--promise furture 同步
http://www.cnblogs.com/haippy/p/3279565.html std::promise 类介绍 promise 对象可以保存某一类型 T 的值,该值可被 future 对象 ...
- YUI Compressor是如何压缩JS代码的?
YUI Compressor 压缩 JavaScript 的内容包括: 移除注释 移除额外的空格 细微优化 标识符替换(Identifier Replacement) YUI Compressor 包 ...
- WCF分布式开发步步为赢(12):WCF事务机制(Transaction)和分布式事务编程
今天我们继续学习WCF分布式开发步步为赢系列的12节:WCF事务机制(Transaction)和分布式事务编程.众所周知,应用系统开发过程中,事务是一个重要的概念.它是保证数据与服务可靠性的重要机制. ...
- The 13th Zhejiang Provincial Collegiate Programming Contest - C
Defuse the Bomb Time Limit: 2 Seconds Memory Limit: 65536 KB The bomb is about to explode! Plea ...
- hive连接数
使用hive分析日志作业很多的时候,需要修改mysql的默认连接数 修改方法 打开/etc/my.cnf文件 在[mysqld] 中添加 max_connections=1000 重启mysql ...
- js和jquery修改背景颜色的区别
html: <HTML> <head> <meta http-equiv="content-type" content="text/html ...
- c++(类继承)示例[仅用于弱弱的博主巩固知识点用哦,不好勿喷]
测试代码: Animals.h: #pragma once #include<string> class Animals { protected: std::string Food; in ...