Array of integers is unimodal, if:

  • it is strictly increasing in the beginning;
  • after that it is constant;
  • after that it is strictly decreasing.

The first block (increasing) and the last block (decreasing) may be absent. It is allowed that both of this blocks are absent.

For example, the following three arrays are unimodal: [5, 7, 11, 11, 2, 1], [4, 4, 2], [7], but the following three are not unimodal:[5, 5, 6, 6, 1], [1, 2, 1, 2], [4, 5, 5, 6].

Write a program that checks if an array is unimodal.

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the array.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1 000) — the elements of the array.

Output

Print "YES" if the given array is unimodal. Otherwise, print "NO".

You can output each letter in any case (upper or lower).

Examples
input
6
1 5 5 5 4 2
output
YES
input
5
10 20 30 20 10
output
YES
input
4
1 2 1 2
output
NO
input
7
3 3 3 3 3 3 3
output
YES
Note

In the first example the array is unimodal, because it is strictly increasing in the beginning (from position 1 to position 2, inclusively), that it is constant (from position 2 to position 4, inclusively) and then it is strictly decreasing (from position 4 to position 6, inclusively).


  题目大意 给定一个数组,判断它是否是单峰的。一个数组是单峰的是指它的最大值出现的位置是连续的,其左侧严格递增,右侧严格递减。

  先找出数组中的最大值,然后while到遇到最大值停止(边判断),然后while把最大值的连续一段水掉,然后再while到数组结尾。

Code

 /**
* Codeforces
* Problem#831A
* Accepted
* Time:15ms
* Memory:2052k
*/
#include <iostream>
#include <cstdio>
#include <ctime>
#include <cmath>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <stack>
#include <cassert>
#ifndef WIN32
#define Auto "%lld"
#else
#define Auto "%I64d"
#endif
using namespace std;
typedef bool boolean;
const signed int inf = (signed)((1u << ) - );
const signed long long llf = (signed long long)((1ull << ) - );
const double eps = 1e-;
const int binary_limit = ;
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
template<typename T>
inline boolean readInteger(T& u){
char x;
int aFlag = ;
while(!isdigit((x = getchar())) && x != '-' && x != -);
if(x == -) {
ungetc(x, stdin);
return false;
}
if(x == '-'){
x = getchar();
aFlag = -;
}
for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
ungetc(x, stdin);
u *= aFlag;
return true;
} int n;
int *a;
int maxv = ; inline void init() {
readInteger(n);
a = new int[(n + )];
for(int i = ; i <= n; i++) {
readInteger(a[i]);
smax(maxv, a[i]);
}
} inline void solve() {
int last;
int i = ;
while(a[i] < maxv) {
if(i != ) {
if(a[i - ] >= a[i]) {
puts("NO");
return;
}
}
i++;
}
while(a[i] == maxv && i <= n) i++;
while(i <= n) {
if(a[i] >= a[i - ]) {
puts("NO");
return;
}
i++;
}
puts("YES");
} int main() {
init();
solve();
return ;
}

Problem A


There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.

You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.

You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.

Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.

Input

The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.

The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.

The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.

Output

Print the text if the same keys were pressed in the second layout.

Examples
input
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
output
HelloVKCup2017
input
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
output
7uduGUDUUDUgudu7

  题目大意 给定字母的映射,然后映射一个字符串,非字母字符保留。

  依题意乱搞即可。

Code

 /**
* Codeforces
* Problem#831B
* Accepted
* Time:15ms
* Memory:2052k
*/
#include <iostream>
#include <cstdio>
#include <ctime>
#include <cmath>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <stack>
#include <cassert>
#ifndef WIN32
#define Auto "%lld"
#else
#define Auto "%I64d"
#endif
using namespace std;
typedef bool boolean;
const signed int inf = (signed)((1u << ) - );
const signed long long llf = (signed long long)((1ull << ) - );
const double eps = 1e-;
const int binary_limit = ;
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
template<typename T>
inline boolean readInteger(T& u){
char x;
int aFlag = ;
while(!isdigit((x = getchar())) && x != '-' && x != -);
if(x == -) {
ungetc(x, stdin);
return false;
}
if(x == '-'){
x = getchar();
aFlag = -;
}
for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
ungetc(x, stdin);
u *= aFlag;
return true;
} int n;
char a[];
char b[];
map<char, char> ctc; const char utl = 'a' - 'A'; inline void init() {
gets(a);
gets(b);
for(int i = ; a[i]; i++) {
ctc[a[i]] = b[i];
ctc[a[i] - utl] = b[i] - utl;
}
} inline void solve() {
gets(a);
for(int i = ; a[i]; i++) {
if((a[i] >= 'a' && a[i] <= 'z') || (a[i] >= 'A' && a[i] <= 'Z')) {
putchar(ctc[a[i]]);
} else {
putchar(a[i]);
}
}
} int main() {
init();
solve();
return ;
}

Problem B

Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem A - B的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) Problem E (Codeforces 828E) - 分块

    Everyone knows that DNA strands consist of nucleotides. There are four types of nucleotides: "A ...

  6. 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 ...

  7. 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: ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. WebSocket.之.基础入门-建立连接

    WebSocket.之.基础入门-建立连接 1. 使用开发工具(STS.Eclipse等)创建web项目.如下图所示,啥东西都没有.一个新的web项目. 2. 创建java类.index.jsp页面. ...

  2. Unity shader学习之菲涅耳反射

    菲涅尔反射(Fresnel reflection),指光线照射物体表面时,一部分发生反射,一部分进入物体内部发生折射或散射,被反射的光和折射光之间存在一定的比率. 2个公式: 1. Schlick 菲 ...

  3. Linq To SQL LEFT OUTER JOIN (Left Join)

    SQL: SELECT [t0].[ProductName], [t1].[TotalPrice] AS [TotalPrice] FROM [Product] AS [t0] LEFT OUTER ...

  4. 原生js实现图片轮播效果

    思路:设置父容器(一定宽度,一定高度,相对定位,子容器超出部分进行隐藏),子容器图片并排(浮动,绝对定位,每次点击进行相应的左或右偏移量) 1.html: <!DOCTYPE html> ...

  5. JQuery基本知识(3)

    JQuery基本知识(3) 一.JQuery拥有可操作HTML元素和属性的强大方法. 1.JQuery DOM操作(DOM文档对象模型) 获取内容的方法: text():设置或返回所选元素的文本内容 ...

  6. Solid Dominoes Tilings (轮廓线dp打表 + 容器)

    第一步先打一个表,就是利用轮廓线DP去打一个没有管有没有分界线组合数量的表 #include<bits/stdc++.h> using namespace std; ; <<; ...

  7. MQTT 发布者订阅者

    添加依赖: <dependency> <groupId>org.eclipse.paho</groupId> <artifactId>org.eclip ...

  8. 转:【专题十二】实现一个简单的FTP服务器

    引言: 休息一个国庆节后好久没有更新文章了,主要是刚开始休息完心态还没有调整过来的, 现在差不多进入状态了, 所以继续和大家分享下网络编程的知识,在本专题中将和大家分享如何自己实现一个简单的FTP服务 ...

  9. 【前端安全】JavaScript防流量劫持

    劫持产生的原因和方式 在网页开发的访问过程中,http是我们主要的访问协议.我们知道http是一种无状态的连接.即没有验证通讯双方的身份,也没有验证信息的完整性,所以很容易受到篡改.运营商就是利用了这 ...

  10. 怎样从外网访问内网Linux系统?

    本地安装了一个Linux系统,只能在局域网内访问到,怎样从外网也能访问到本地的Linux系统呢?本文将介绍具体的实现步骤. 1. 准备工作 1.1 启动Linux系统 默认Linux系统ssh服务端端 ...