Sort the Array CodeForces

Sort the Array CodeForces - 451B

题目描述
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 ≤ 10^5) — the size of array a.

The second line contains n distinct space-separated integers: a[1], a[2], …, a[n] (1 ≤ a[i] ≤ 10^9).

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.

Example

Input
3
3 2 1
Output
yes
1 3

Input
4
2 1 3 4
Output
yes
1 2

Input
4
3 1 2 4
Output
no

Input
2
1 2
Output
yes
1 1

Note
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].

大致题意
给你一个数组，下标从1开始，问你是否可以反转一段区间内的数字，使得这个数组单调递增。

思路，遍历整个数组a，找到第一个a[i]>a[i+1]的位置，将i值赋给x，然后从x开始遍历，找到第一个a[i]< a[i+1]的位置，将i赋给y，用reverse （）进行（x，y）区间内的反转，然后再遍历整个数组看是否单调递增。

代码如下

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<vector>#include<cmath>#include<string>#define LL long long  const int  Max =1e6+5;using namespace std;int a[100005];int main(){    int n,x=1,y=1,flag=1;    cin>>n;    for(int i=1;i<=n;i++)     cin>>a[i];     for(int i=1;i<n;i++)     if(a[i]>a[i+1]) {        x=i;        break;     }     int i;     for(i=x;i<n;i++)     if(a[i]<a[i+1]){        y=i;        break;     }     if(i==n) y=n;     reverse(a+x,a+y+1);     for(int i=1;i<n;i++)     if(a[i]>a[i+1]){        flag=0;        break;     }     if(flag==1){        cout<<"yes"<<endl;        cout<<x<<' '<<y;     }     else      cout<<"no"<<endl;   return 0;}