CodeForces 451B
Sort the Array
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Description
Being a programmer, you like arrays a lot. For your birthday, your friends have given you an array a consisting of n distinct integers.
Unfortunately, the size of a is too small. You want a bigger array! Your friends agree to give you a bigger array, but only if you are able to answer the following question correctly: is it possible to sort the array a (in increasing order) by reversing exactly one segment of a? See definitions of segment and reversing in the notes.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 105) — the size of array a.
The second line contains n distinct space-separated integers: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109).
Output
Print "yes" or "no" (without quotes), depending on the answer.
If your answer is "yes", then also print two space-separated integers denoting start and end (start must not be greater than end) indices of the segment to be reversed. If there are multiple ways of selecting these indices, print any of them.
Sample Input
3
3 2 1
yes
1 3
4
2 1 3 4
yes
1 2
4
3 1 2 4
no
2
1 2
yes
1 1
//2016.8.2
#include<iostream>
#include<cstdio>
#include<algorithm> using namespace std; int main()
{
long long arr_src[], arr_dest[];
int l, r, n;
while(cin >> n)
{
r = l = ;
for(int i = ; i <= n; i++)
{
scanf("%lld", &arr_src[i]);
arr_dest[i] = arr_src[i];
}
sort(arr_dest+, arr_dest++n);
for(int i = ; i <= n; i++)
if(arr_src[i] != arr_dest[i])
{
l = i;
break;
}
for(int i = n; i >= ; i--)
if(arr_src[i]!=arr_dest[i])
{
r = i;
break;
}
bool fg = true;
reverse(arr_dest+l, arr_dest++r);
for(int i = ; i <= n; i++)
{
if(arr_src[i] != arr_dest[i])
{
fg = false;
break;
}
}
if(fg)printf("yes\n%d %d\n", l, r);
else printf("no\n");
}
return ;
}
CodeForces 451B的更多相关文章
- CF Codeforces Round #258 (Div. 2) B (451B)
题意:找出一段逆序! 预存a[]数组到b[]数组.将b排序,然后前后找不同找到区间[l,r],然后推断[l,r]是否逆序就能够了!.当然还得特判本身就是顺序的!! ! AC代码例如以下: #inclu ...
- Codeforces Round #258 (Div. 2) 小结
A. Game With Sticks (451A) 水题一道,事实上无论你选取哪一个交叉点,结果都是行数列数都减一,那如今就是谁先减到行.列有一个为0,那么谁就赢了.因为Akshat先选,因此假设行 ...
- python爬虫学习(5) —— 扒一下codeforces题面
上一次我们拿学校的URP做了个小小的demo.... 其实我们还可以把每个学生的证件照爬下来做成一个证件照校花校草评比 另外也可以写一个物理实验自动选课... 但是出于多种原因,,还是绕开这些敏感话题 ...
- 【Codeforces 738D】Sea Battle(贪心)
http://codeforces.com/contest/738/problem/D Galya is playing one-dimensional Sea Battle on a 1 × n g ...
- 【Codeforces 738C】Road to Cinema
http://codeforces.com/contest/738/problem/C Vasya is currently at a car rental service, and he wants ...
- 【Codeforces 738A】Interview with Oleg
http://codeforces.com/contest/738/problem/A Polycarp has interviewed Oleg and has written the interv ...
- CodeForces - 662A Gambling Nim
http://codeforces.com/problemset/problem/662/A 题目大意: 给定n(n <= 500000)张卡片,每张卡片的两个面都写有数字,每个面都有0.5的概 ...
- CodeForces - 274B Zero Tree
http://codeforces.com/problemset/problem/274/B 题目大意: 给定你一颗树,每个点上有权值. 现在你每次取出这颗树的一颗子树(即点集和边集均是原图的子集的连 ...
- CodeForces - 261B Maxim and Restaurant
http://codeforces.com/problemset/problem/261/B 题目大意:给定n个数a1-an(n<=50,ai<=50),随机打乱后,记Si=a1+a2+a ...
随机推荐
- LintCode 11 二叉查找树的搜索区间
题目链接:http://www.lintcode.com/zh-cn/problem/search-range-in-binary-search-tree/ 1.描述 给定两个值 k1 和 k2(k1 ...
- Java网络通信——XML和JSON
XML(Extensible Markup Language) 定义:一种可扩展的标记性语言 XML有丰富的编码工具,比如Dom4j.JDom等. JSON(JavaScript Object Not ...
- 浅谈Linux集群
集群听起来好像就是一个很高端很的技术,其实不是的,那么集群其实就是一堆计算机的集合,给用户提供同一个服务的一组计算机,就称之为集群,对于用户而言好像就是一台计算机提供的服务,集群主要分为三大类, ...
- Apache Commons工具集简介(转)
此文为转帖,原帖地址:http://zhoualine.iteye.com/blog/1770014
- python之路: 基础篇
)或>>> name = ) #按照占位符的顺序):] #下标识从0开始的 wulaoer >>> print name[:] # ...
- Struts2---OGNL表达式和EL表达式
在action里放入actioncontext的变量值 ActionContext.getContext().put("forumList", forumList); 在jsp里如 ...
- angularJS 系列(三)- 自定义 Service
参考:http://viralpatel.net/blogs/angularjs-service-factory-tutorial/ https://www.pluralsight.com/blog/ ...
- jquery中如何以逗号分割字符串_百度知道
body{ font-family: "Microsoft YaHei UI","Microsoft YaHei",SimSun,"Segoe UI& ...
- GoEasy消息推送
1. 从GoEasy获取appkey appkey是验证用户的有效性的唯一标识. 注册账号. GoEasy官网:https://goeasy.io 用注册好的账号登录到GoEasy的后台管理系统,创建 ...
- php smarty
摘自:http://linux.chinaitlab.com/PHP/38324.html 刚开始接触模版引擎的 PHP 设计师,听到 Smarty 时,都会觉得很难.其实笔者也不例外,碰都不敢碰一 ...