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$d\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d} \right \rceil - \sum ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem D (Codeforces 831D) - 贪心 - 二分答案 - 动态规划
There are n people and k keys on a straight line. Every person wants to get to the office which is l ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem E (Codeforces 831E) - 线段树 - 树状数组
Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this int ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem C (Codeforces 831C) - 暴力 - 二分法
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem A - B
Array of integers is unimodal, if: it is strictly increasing in the beginning; after that it is cons ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) E. Cards Sorting 树状数组
E. Cards Sorting time limit per test 1 second memory limit per test 256 megabytes input standard inp ...
随机推荐
- 【bzoj1018】[SHOI2008]堵塞的交通traffic 线段树区间合并+STL-set
题目描述 给出一张2*n的网格图,初始每条边都是不连通的.多次改变一条边的连通性或询问两个点是否连通. 输入 第一行只有一个整数C,表示网格的列数.接下来若干行,每行为一条交通信息,以单独的一行“Ex ...
- jQuery插件jquery.fullPage.js
简介如今我们经常能看到全屏网站,尤其是国外网站.这些网站用几幅很大的图片或者色块做背景,再添加一些简单的内容,显得格外的高端大气上档次,比如 iPone 5C 的介绍页面.QQ浏览器的官方网站.百度史 ...
- [洛谷P1892]团伙
题目大意:有n个人,关系为:朋友的朋友是朋友,敌人的敌人是朋友.如果是朋友就在一个团队内,是敌人就不在,现在给出一关系,问最多有多少团伙.题解:并查集,建反集,如果是朋友,就把他们的并查集合并:如果是 ...
- Flash by sshockwave [树dp]
题目 给定一棵树,每个点有一个活动时间,长度为正整数$t_i$ 你需要安排每个点的活动时间什么时候开始什么时候结束,并且满足:任何一个时刻没有两个相邻的点都在活动 开始时刻为0,在以上条件下最小化所有 ...
- [Leetcode] remove element 删除元素
Given an array and a value, remove all instances of that value in place and return the new length. T ...
- picks loves segment tree I
picks loves segment tree I 题目背景 来源: \(\text {2018 WC Segment Tree Beats}\) 原作者: \(\text {C_SUNSHINE} ...
- hadoop SecondNamenode 详解
SecondNamenode名字看起来很象是对第二个Namenode,要么与Namenode一样同时对外提供服务,要么相当于Namenode的HA. 真正的了解了SecondNamenode以后,才发 ...
- NOIP2016愤怒的小鸟 [状压dp]
愤怒的小鸟 题目描述 Kiana 最近沉迷于一款神奇的游戏无法自拔. 简单来说,这款游戏是在一个平面上进行的. 有一架弹弓位于 (0,0) 处,每次 Kiana 可以用它向第一象限发射一只红色的小鸟, ...
- 解读python小练习
1.新建一个函数,判断是不是int 类型,并测试,不是抛出错误def adder(x, y):"""Return x + y if they are both integ ...
- python异常之with
1.基本语法 with expression [as target]: with_body 参数说明: expression:是一个需要执行的表达式: target:是一个变量或者元组,存储的是exp ...