CodeForces 451B Sort the Array

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 test(s)

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].

 

题意：给你一个长度不超过10^5的数组，数组中的每个元素互不相同。

问是否能翻转数组的某个区间，使得数组形成一个有序的升序序列。

 

两次循环，第一次从左到右找出比前一个数小的数字，记录下标

第二次从右到左找出比后一个数大的数字，记录下标

反转两个下标直接的数字

检验得到的数组序列是否为递增

 1 #include<stdio.h>

 2 #include<algorithm>

 3 #define maxn 100005

 4 

 5 using namespace std;

 6 int a[maxn];

 7 int main()

 8 {

 9     int n, i, j;

10     while(scanf("%d", &n) != EOF){

11         int x = 1, y = 1;

12         for(i = 1; i <= n; i++)

13             scanf("%d",&a[i]);

14 

15         for(i = 1; i < n; i++)

16         if(a[i] > a[i+1]){

17             x = i; break;

18         }

19         for(i = n; i >= 1; i--)

20         if(a[i] < a[i-1]){

21             y = i; break;

22         }

23 

24         reverse(a+x, a+y+1);

25         int flag=0;

26         for(i = 1; i < n; i++)

27             if(a[i] > a[i+1]){

28                 flag = 1; break;

29             }

30         if(flag)

31             printf("no\n");

32         else

33             printf("yes\n%d %d\n", x, y);

34     }

35     return 0;

36 }