CF - 451B. Sort the Array 排序+模拟

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本文介绍了一个编程挑战，任务是判断是否可以通过反转数组中的一段子区间来实现数组的递增排序。文章提供了解决问题的思路和AC代码示例。

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1.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 ≤ 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.

Examples

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].

2.题意概述：

给你一段数，要你判断是否能够通过逆序一段字区间使得数严格不递减

3.解题思路：

1、从左到右遍历，第一次出现比前一个数小的下标；
2、然后从右往左遍历，第一次出现比后一个数小的下标；
3、将这段区间逆序，然后判断整个序列是否单调递增

4.AC代码：

#include <stdio.h>
#include <algorithm>
#define maxn 100100
using namespace std;
int a[maxn];
int main()
{
	int n;
	while (scanf("%d", &n) != EOF)
	{
		int flag = 0, x = 0, y = 0;
		for (int i = 0; i < n; i++)
			scanf("%d", &a[i]);
		for (int i = 0; i < n - 1; i++)
			if (a[i] > a[i + 1])
			{
				x = i;
				break;
			}
		for (int i = n - 1; i > 0; i--)
			if (a[i] < a[i - 1])
			{
				y = i;
				break;
			}
		reverse(a + x, a + y + 1);
		for (int i = 0; i < n - 1; i++)
		{
			if (a[i] > a[i + 1])
			{
				flag = 1;
				break;
			}
		}
		if (flag)
			puts("no");
		else
		{
			printf("yes\n%d %d\n", x + 1, y + 1);
		}


	}
	return 0;


}