CF--思维练习--CodeForces - 221C-H - Little Elephant and Problem (思维)
ACM思维题训练集合
The Little Elephant has got a problem — somebody has been touching his sorted by non-decreasing array a of length n and possibly swapped some elements of the array.
The Little Elephant doesn't want to call the police until he understands if he could have accidentally changed the array himself. He thinks that he could have accidentally changed array a, only if array a can be sorted in no more than one operation of swapping elements (not necessarily adjacent). That is, the Little Elephant could have accidentally swapped some two elements.
Help the Little Elephant, determine if he could have accidentally changed the array a, sorted by non-decreasing, himself.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the size of array a. The next line contains n positive integers, separated by single spaces and not exceeding 109, — array a.
Note that the elements of the array are not necessarily distinct numbers.
Output
In a single line print "YES" (without the quotes) if the Little Elephant could have accidentally changed the array himself, and "NO" (without the quotes) otherwise.
Examples
Input
2
1 2
Output
YES
Input
3
3 2 1
Output
YES
Input
4
4 3 2 1
Output
NO
Note
In the first sample the array has already been sorted, so to sort it, we need 0 swap operations, that is not more than 1. Thus, the answer is "YES".
In the second sample we can sort the array if we swap elements 1 and 3, so we need 1 swap operation to sort the array. Thus, the answer is "YES".
In the third sample we can't sort the array in more than one swap operation, so the answer is "NO".
题目中说原来的数列非递减,也就是非严格递增,显然原数列易得,排遍序即可,如果发生了一次交换至多有两个不一样。
#include <bits/stdc++.h>
using namespace std;
const int maxn=100055;
int a[maxn];
int b[maxn];
int main()
{
int n;
cin>>n;
for(int i=0;i<n;i++)
{
cin>>a[i];
b[i]=a[i];
}
sort(b,b+n);
int cnt=0;
for(int i=0;i<n;i++)
{
if(a[i]!=b[i]) cnt++;
if(cnt==3) break;
}
if(cnt>2) cout<<"NO"<<endl;
else cout<<"YES"<<endl;
}
CF--思维练习--CodeForces - 221C-H - Little Elephant and Problem (思维)的更多相关文章
- Codeforces 221 C. Little Elephant and Problem
C. Little Elephant and Problem time limit per test 2 seconds memory limit per test 256 megabytes inp ...
- Codeforces 221d D. Little Elephant and Array
二次联通门 : Codeforces 221d D. Little Elephant and Array /* Codeforces 221d D. Little Elephant and Array ...
- Educational Codeforces Round 40 F. Runner's Problem
Educational Codeforces Round 40 F. Runner's Problem 题意: 给一个$ 3 * m \(的矩阵,问从\)(2,1)$ 出发 走到 \((2,m)\) ...
- 2019~2020icpc亚洲区域赛徐州站H. Yuuki and a problem
2019~2020icpc亚洲区域赛徐州站H. Yuuki and a problem 题意: 给定一个长度为\(n\)的序列,有两种操作: 1:单点修改. 2:查询区间\([L,R]\)范围内所有子 ...
- CF思维联系--CodeForces - 218C E - Ice Skating (并查集)
题目地址:24道CF的DIv2 CD题有兴趣可以做一下. ACM思维题训练集合 Bajtek is learning to skate on ice. He's a beginner, so his ...
- CF思维联系– CodeForces - 991C Candies(二分)
ACM思维题训练集合 After passing a test, Vasya got himself a box of n candies. He decided to eat an equal am ...
- CF思维联系–CodeForces - 225C. Barcode(二路动态规划)
ACM思维题训练集合 Desciption You've got an n × m pixel picture. Each pixel can be white or black. Your task ...
- CF思维联系–CodeForces -224C - Bracket Sequence
ACM思维题训练集合 A bracket sequence is a string, containing only characters "(", ")", ...
- CF思维联系–CodeForces - 223 C Partial Sums(组合数学的先线性递推)
ACM思维题训练集合 You've got an array a, consisting of n integers. The array elements are indexed from 1 to ...
随机推荐
- java中OOM错误解析(面试可以聊的东西)
嗯,生活加油鸭.... 实习中遇到OOM错误 GC overhead limit exceeded 问题,所以整理一下OOM异常问题: 先看一下“阿里的开发手册”对OOM的描述: OOM,全称“Out ...
- js实现表单的隔行换色、鼠标高亮出来等相关内容以及相关事件的作用
主要是使用的onload().onmouseover和onmouseout的相关应用,满足此次的相关操作. 具体的相关的两个代码如下: <!DOCTYPE html> <html&g ...
- @suppressWarnings("unchecked") java 中是什么意思 (一般放dao查询方法上)
J2SE 提供的最后一个批注是 @SuppressWarnings.该批注的作用是给编译器一条指令,告诉它对被批注的代码元素内部的某些警告保持静默. 一点背景:J2SE 5.0 为 Java 语言增加 ...
- 在IDEA中搭建Java源码学习环境并上传到GitHub上
打开IDEA新建一个项目 创建一个最简单的Java项目即可 在项目命名填写该项目的名称,我这里写的项目名为Java_Source_Study 点击Finished,然后在项目的src目录下新建源码文件 ...
- GeoGebra重复手段实现
1.自定义工具部分可以在网上搜一些别人做的工具,主要是把自己经常做的一些任务做成工具,减少重复过程 2.列表部分的简单操作如图所示,实现对三个点的多项式拟合 3.通过序列指令格式可以做一个好玩的效果, ...
- Salesforce学习 | 系统管理员Admin如何添加用户
作为世界排名第一的CRM云计算软件,不管的是500强还是中小企业,越来越多的公司都选择使用Salesforce来分享客户信息,管理和开发具有更高收益的客户关系.Salesforce Administr ...
- Delphi 文件操作(4)Reset
procedure Reset(var F [: File; RecSize: Word ] ); { 作用: 对于文本文件,Reset过程将以只读方式打开文件,对于类型文件和无类型文件, ...
- SpringBoot连接Redis服务出现DENIED Redis is running in protected mode because protected mode is enabled
修改redis.conf,yes 改为 no
- flutter和react native如何选择
[关于性能]跨平台开发第一个考虑的就是性能问题RN的效率由于是将View编译成了原生View,所以效率上要比基于Cordova的HTML5高很多,但是它也有效率问题,RN的渲染机制是基于前端框架的考虑 ...
- 词向量模型word2vector详解
目录 前言 1.背景知识 1.1.词向量 1.2.one-hot模型 1.3.word2vec模型 1.3.1.单个单词到单个单词的例子 1.3.2.单个单词到单个单词的推导 2.CBOW模型 3.s ...