B. Array Sharpening

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

You’re given an array a1,…,an of n non-negative integers.

Let’s call it sharpened if and only if there exists an integer 1≤k≤n such that a1<a2<…ak+1>…>an. In particular, any strictly increasing or strictly decreasing array is sharpened. For example:

The arrays [4], [0,1], [12,10,8] and [3,11,15,9,7,4] are sharpened;

The arrays [2,8,2,8,6,5], [0,1,1,0] and [2,5,6,9,8,8] are not sharpened.

You can do the following operation as many times as you want: choose any strictly positive element of the array, and decrease it by one. Formally, you can choose any i (1≤i≤n) such that ai>0 and assign ai:=ai−1.

Tell if it’s possible to make the given array sharpened using some number (possibly zero) of these operations.

Input

The input consists of multiple test cases. The first line contains a single integer t (1≤t≤15 000) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (1≤n≤3⋅105).

The second line of each test case contains a sequence of n non-negative integers a1,…,an (0≤ai≤109).

It is guaranteed that the sum of n over all test cases does not exceed 3⋅105.

Output

For each test case, output a single line containing “Yes” (without quotes) if it’s possible to make the given array sharpened using the described operations, or “No” (without quotes) otherwise.

Example

inputCopy

10

1

248618

3

12 10 8

6

100 11 15 9 7 8

4

0 1 1 0

2

0 0

2

0 1

2

1 0

2

1 1

3

0 1 0

3

1 0 1

outputCopy

Yes

Yes

Yes

No

No

Yes

Yes

Yes

Yes

No

Note

In the first and the second test case of the first test, the given array is already sharpened.

In the third test case of the first test, we can transform the array into [3,11,15,9,7,4] (decrease the first element 97 times and decrease the last element 4 times). It is sharpened because 3<11<15 and 15>9>7>4.

In the fourth test case of the first test, it’s impossible to make the given array sharpened.

这个题考虑不能构成的情况,就是从这数左边不能构成,右边也不能构成。然后左右都可以的时候,中间两个可能相等不能构成左右。

#include <bits/stdc++.h>
using namespace std;
int a[300010];
int main()
{
int T;
cin >> T;
while (T--)
{
int n;
cin>>n;
bool bk = true;
for (int i = 1; i <= n; i++)
cin>>a[i] ;
for (int i = 1; i <= n; i++)
if (a[i] < min(i - 1, n - i))
{
bk = false;
break;
}
if (n % 2 == 0 && max(a[n / 2], a[n / 2 + 1]) < n / 2)
{
bk = false;
}
if (bk == false)
puts("No");
else
puts("Yes");
}
return 0;
}

Codeforces 1291 Round #616 (Div. 2) B的更多相关文章

  1. Codeforces 1291 Round #616 (Div. 2) C. Mind Control(超级详细)

    C. Mind Control You and your n−1 friends have found an array of integers a1,a2,-,an. You have decide ...

  2. Codeforces Beta Round #80 (Div. 2 Only)【ABCD】

    Codeforces Beta Round #80 (Div. 2 Only) A Blackjack1 题意 一共52张扑克,A代表1或者11,2-10表示自己的数字,其他都表示10 现在你已经有一 ...

  3. Codeforces Beta Round #83 (Div. 1 Only)题解【ABCD】

    Codeforces Beta Round #83 (Div. 1 Only) A. Dorm Water Supply 题意 给你一个n点m边的图,保证每个点的入度和出度最多为1 如果这个点入度为0 ...

  4. Codeforces Beta Round #79 (Div. 2 Only)

    Codeforces Beta Round #79 (Div. 2 Only) http://codeforces.com/contest/102 A #include<bits/stdc++. ...

  5. Codeforces Beta Round #77 (Div. 2 Only)

    Codeforces Beta Round #77 (Div. 2 Only) http://codeforces.com/contest/96 A #include<bits/stdc++.h ...

  6. Codeforces Beta Round #76 (Div. 2 Only)

    Codeforces Beta Round #76 (Div. 2 Only) http://codeforces.com/contest/94 A #include<bits/stdc++.h ...

  7. Codeforces Beta Round #75 (Div. 2 Only)

    Codeforces Beta Round #75 (Div. 2 Only) http://codeforces.com/contest/92 A #include<iostream> ...

  8. Codeforces Beta Round #74 (Div. 2 Only)

    Codeforces Beta Round #74 (Div. 2 Only) http://codeforces.com/contest/90 A #include<iostream> ...

  9. Codeforces Beta Round #73 (Div. 2 Only)

    Codeforces Beta Round #73 (Div. 2 Only) http://codeforces.com/contest/88 A 模拟 #include<bits/stdc+ ...

随机推荐

  1. WordPress文章阅读量统计和显示(非插件, 刷新页面不累加)

    本文已同步到专业技术网站 www.sufaith.com, 该网站专注于前后端开发技术与经验分享, 包含Web开发.Nodejs.Python.Linux.IT资讯等板块. WordPress文章阅读 ...

  2. JS 浏览器BOM-->open() 方法

    1.定义和用法 open() 方法用于打开一个新的浏览器窗口或查找一个已命名的窗口. 语法: window.open(URL,name,specs,replace) 参数: URL:打开指定的页面的U ...

  3. 【Selenium04篇】python+selenium实现Web自动化:文件上传,Cookie操作,调用 JavaScript,窗口截图

    一.前言 最近问我自动化的人确实有点多,个人突发奇想:想从0开始讲解python+selenium实现Web自动化测试,请关注博客持续更新! 这是python+selenium实现Web自动化第四篇博 ...

  4. AJ学IOS(45)之常用的小功能比如打电话、打开网址、发邮件、发短信打开其他应用。

    AJ分享,必须精品 简介 iOS中的很多小功能都是非常简单的,几行代码就搞定了,比如打电话.打开网址.发邮件.发短信.打开其他应用等. 打电话 方法1 最简单最直接的方式:直接跳到拨号界面 NSURL ...

  5. 使用python pip安装工具组件包:出现 requests File “<stdin>",line 1 pip install xxx ^ SyntaxError:invalid syntax

    最近想要试试python ,软件安装完成了,但是,import 组件包时,出了问题,一直不得解:安装pycharm 工具感觉麻烦,不想安装那些,只想单纯使用python . 问题复现: 1.Windo ...

  6. python基础之函数详解

    Python基础之函数详解 目录 Python基础之函数详解 一.函数的定义 二.函数的调用 三.函数返回值 四.函数的参数 4.1 位置参数 4.2 关键字参数 实参:位置实参和关键字参数的混合使用 ...

  7. L8梯度消失、梯度爆炸

    houseprices数据下载: 链接:https://pan.baidu.com/s/1-szkkAALzzJJmCLlJ1aXGQ 提取码:9n9k 梯度消失.梯度爆炸以及Kaggle房价预测 代 ...

  8. ASE课程总结 by 张葳

    本期ASE课程分为两个阶段,第一阶段的personal project与第二阶段的team project,其中,第一阶段旨在锻炼我们个人的问题解决能力和编程能力,第二阶段则锻炼主要我们的管理能力,合 ...

  9. 原创Pig0.16.0安装搭建

    tar -zxvf pig-0.16.0.tar.gz -C ~   vi ~/.bash_profile export PIG_HOME=/home/hadoop/pig-0.16.0 export ...

  10. Linux 平台 安装 Composer

    1.检查是否安装 composer --version 2.下载安装 php -r "copy('https://install.phpcomposer.com/installer', 'c ...