Codeforces 221 C. Little Elephant and Problem
2 seconds
256 megabytes
standard input
standard output
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.
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.
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.
2
1 2
YES
3
3 2 1
YES
4
4 3 2 1
NO
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".
题意:给出一个长为n的序列,问是否能够交换至多一次,使序列非降
#include<cstdio>
#include<algorithm>
using namespace std;
int n;
int a[],b[];
int main()
{
scanf("%d",&n);
for(int i=;i<=n;i++)
{
scanf("%d",&a[i]);
b[i]=a[i];
}
sort(b+,b+n+);
int f=;
for(int i=;i<=n;i++)
if(a[i]!=b[i])
{
if(f<) f++;
else { puts("NO"); return ; }
}
puts("YES");
}
Codeforces 221 C. Little Elephant and Problem的更多相关文章
- Codeforces 221 B. Little Elephant and Numbers
B. Little Elephant and Numbers time limit per test 2 seconds memory limit per test 256 megabytes inp ...
- Codeforces 221 E. Little Elephant and Shifts
E. Little Elephant and Shifts time limit per test 2 seconds memory limit per test 256 megabytes inpu ...
- Codeforces 221 D. Little Elephant and Array
D. Little Elephant and Array time limit per test 4 seconds memory limit per test 256 megabytes input ...
- Codeforces 221 A. Little Elephant and Function
A. Little Elephant and Function time limit per test 2 seconds memory limit per test 256 megabytes in ...
- 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 \((2,m)\) ...
- 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 ...
- 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-decr ...
- AC日记——Little Elephant and Problem codeforces 221c
221C 思路: 水题: 代码: #include <cstdio> #include <cstring> #include <iostream> #include ...
- Codeforces Beta Round #17 A - Noldbach problem 暴力
A - Noldbach problem 题面链接 http://codeforces.com/contest/17/problem/A 题面 Nick is interested in prime ...
随机推荐
- PSP1130
PSP时间图: 类型 任务 开始时间 结束时间 净时间 中断时间 日期 开会 开会 16:17 16:50 33 0 20171027 开会 开会 17:00 17:22 22 0 20171028 ...
- 2018软工实践—Alpha冲刺(5)
队名 火箭少男100 组长博客 林燊大哥 作业博客 Alpha 冲鸭鸭鸭鸭鸭! 成员冲刺阶段情况 林燊(组长) 过去两天完成了哪些任务 协调各成员之间的工作 协助测试的进行 测试项目运行的服务器环境 ...
- 404 Note Found 现场编程
目录 组员职责分工 github 的提交日志截图 程序运行截图 程序运行环境 GUI界面 基础功能实现 运行视频 LCG算法 过滤(降权)算法 算法思路 红黑树 附加功能一 背景 实现 附加功能二(迭 ...
- MySQL 日志功能详解
MySQL日志分类 1:查询日志 :query log 2:慢查询日志:slow_query_log 查询执行时长超过指定时长的查询操作所记录日志 3:错误日志:error log ...
- 1106C程序语法树
- 软工实践团队展示——WorldElite
软工实践团队展示--WorldElite 本次作业链接 团队成员 031602636许舒玲(组长) 031602237吴杰婷 031602634吴志鸿 081600107傅滨 031602220雷博浩 ...
- 【第五周】四则运算GUI
这次这个简陋的程序终于发布了,其实发布很简单(在windows平台),因为使用的是vs2008+qt4.7的组合,在微软自家平台上用一用还是很方便的,只需要在release编译生成的exe文件,加上几 ...
- Node.js系列——(4)优势及场景
背景 之前几篇系列文章简单介绍了node.js的安装配置及基本操作: Node.js系列--(1)安装配置与基本使用 Node.js系列--(2)发起get/post请求 Node.js系列--(3) ...
- C++11 锁 lock
转自:https://www.cnblogs.com/diegodu/p/7099300.html 互斥(Mutex: Mutual Exclusion) 下面的代码中两个线程连续的往int_set中 ...
- cat命令和EOF标识输出shell到文件
在某些场合,可能我们需要在脚本中生成一个临时文件,然后把该文件作为最终文件放入目录中.(可参考ntop.spec文件)这样有几个好处,其中之一就是临时文件不是唯一的,可以通过变量赋值,也可根据不同的判 ...