Codeforces Round #258 (Div. 2/B)/Codeforces451B_Sort the Array
解题报告
http://blog.csdn.net/juncoder/article/details/38102391
对于给定的数组,取对数组中的一段进行翻转,问翻转后是否是递增有序的。
思路:
仅仅要找到最初递减的区域,记录区域内最大和最小的值,和区间位置。
然后把最大值与区间的下一个元素对照,最小值与区间上一个元素对照。
这样还不够,可能会出现两个或两个以上的递减区间,这样的情况直接pass,由于仅仅能翻转一次。
#include <iostream>
#include <cstdio>
#include <cstring>
#define inf 999999999999
#define LL long long
#define swap(x,y,t) ((t)=(x),(x)=(y),(y)=(t))
using namespace std;
LL num[100010];
int main()
{
int n,i,j;
while(cin>>n)
{
for(i=1; i<=n; i++)
cin>>num[i];
for(i=1; i<=n; i++)
{
if(num[i+1]<num[i])
break;
}
num[n+1]=inf;
if(i==n+1)
{
printf("1 1\n");
continue;
}
int l,r;
LL maxx=0,minn=inf;
int t=i;
for(; i<=n; i++)
{
if(maxx<num[i])
{
maxx=num[i];
r=i;
}
if(minn>num[i])
{
l=i;
minn=num[i];
}
if(num[i+1]>num[i])
break;
}
int u=i;
int q=0;
for(; i<=n; i++)
{
if(num[i+1]<num[i])
{
q=1;
printf("no\n");
break;
}
}
if(!q)
if(num[u+1]>=maxx&&num[t-1]<minn)
{
printf("yes\n");
if(l<r)
swap(l,r,q);
printf("%d %d\n",r,l);
}
else printf("no\n");
}
return 0;
}
1 second
256 megabytes
standard input
standard output
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.
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).
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.
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
Sample 1. You can reverse the entire array to get [1, 2, 3], which is sorted.
Sample 3. No segment can be reversed such that the array will be sorted.
Definitions
A segment [l, r] of array a is the sequence a[l], a[l + 1], ..., a[r].
If you have an array a of size n and you reverse
its segment [l, r], the array will become:
a[1], a[2], ..., a[l - 2], a[l - 1], a[r], a[r - 1], ..., a[l + 1], a[l], a[r + 1], a[r + 2], ..., a[n - 1], a[n].
Codeforces Round #258 (Div. 2/B)/Codeforces451B_Sort the Array的更多相关文章
- Codeforces Round #258 (Div. 2) B. Sort the Array
题目链接:http://codeforces.com/contest/451/problem/B 思路:首先找下降段的个数,假设下降段是大于等于2的,那么就直接输出no,假设下降段的个数为1,那么就把 ...
- Codeforces Round #258 (Div. 2) B. Sort the Array(简单题)
题目链接:http://codeforces.com/contest/451/problem/B --------------------------------------------------- ...
- Codeforces Round #258 (Div. 2)——B. Sort the Array
B. Sort the Array time limit per test 1 second memory limit per test 256 megabytes input standard in ...
- Codeforces Round #258 (Div. 2)[ABCD]
Codeforces Round #258 (Div. 2)[ABCD] ACM 题目地址:Codeforces Round #258 (Div. 2) A - Game With Sticks 题意 ...
- 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 ...
- Codeforces Round #258 (Div. 2) E. Devu and Flowers 容斥
E. Devu and Flowers 题目连接: http://codeforces.com/contest/451/problem/E Description Devu wants to deco ...
- Codeforces Round #258 (Div. 2) D. Count Good Substrings 水题
D. Count Good Substrings 题目连接: http://codeforces.com/contest/451/problem/D Description We call a str ...
- Codeforces Round #258 (Div. 2) C. Predict Outcome of the Game 水题
C. Predict Outcome of the Game 题目连接: http://codeforces.com/contest/451/problem/C Description There a ...
- Codeforces Round #258 (Div. 2) . Sort the Array 贪心
B. Sort the Array 题目连接: http://codeforces.com/contest/451/problem/B Description Being a programmer, ...
随机推荐
- POJ3268 Silver Cow Party(dijkstra+矩阵转置)
Silver Cow Party Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 15156 Accepted: 6843 ...
- MAC 下cocos2d-x lua 使用dragonbones的方法
项目使用db,网上查了半天全是vs和android的流程,没查到有mac的.这里记录一下. quick-cocos-x下的使用方法: a. 将dragonbones(放入ucocos2d_libs中) ...
- 使用Comparable接口的小例子
代码: public class Student implements Comparable<Student> { private int id; private String name; ...
- zb的生日(暴搜dfs)
zb的生日 时间限制:3000 ms | 内存限制:65535 KB 难度:2 描述 今天是阴历七月初五,acm队员zb的生日.zb正在和C小加.never在武汉集训.他想给这两位兄弟买点什么 ...
- Ice_cream's world I
Ice_cream's world I Time Limit : 3000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) ...
- myeclipse集成weblogicserver
今天为了学一下JMS的东东, 不得不安装个weblogicserver, 下面是详细的安装步骤: 1. 首先去官网下载一个weblogic: 下载地址: http://download.oracle. ...
- java中String的用法
String的用法很活跃,也用到的很多.可以根据自己的需要查询API.这里只有concat和substring,indexof的用法 class TestString { public static ...
- angular controller之间通信方式
对于日常开发中,难免会有controller之间通信需求.对于controller之间通信有两种方式可以做到. 用 Angular 进行开发,基本上都会遇到 Controller 之间通信的问题,本文 ...
- 浅谈postMessage多页面监听事件
最近做了一个Echarts和Highcharts多图多页面连动的效果,就用到postMessage 如下介绍: 最开始在最外围的页面也就是所有页面的父级页面添加postMessage监听事件以便监听下 ...
- listbox修改字体大小
listBox1.Font = new Font(this.Font.FontFamily, 14);