CF258，div-2，B，排序问题

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本文探讨如何通过反转数组中的一段子区间来判断原始数组是否能够被排序成递增顺序，并提供了一个实现该功能的算法实例。包括输入输出规范、代码解析与运行示例。

原题 http://codeforces.com/contest/451/problem/B

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].

题目大意。给你一串数字，其中有些是逆序的，问如果改变这个逆序，这串数字是不是按增长的顺序排列

//题目看错wa了两次。数组开的太小re了一次。。。。
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <set>
#include <map>
using namespace std;
#define N 111111
int a[N];

int main(){
	int n,i;
	int c,d,flag1,flag2,flag3;

	while(~scanf("%d",&n)){
		for(i=0;i<n;i++){
			scanf("%d",&a[i]);
		}
		flag1 = 2;
		flag2 = 2;
		flag3 = 2;
		for(i=0;i<n-1;i++){
			if(a[i+1] < a[i]){
				c = i;
				flag1 = 1;
				//flag2 = 1;
				break;
			}
		}
		for(i=n-1;i>0;i--){
			if(a[i-1] > a[i]){
				d = i;
				flag2 = 1;
				break;
			}
		}
		if(flag2!=1 && flag2!=1){
			printf("yes\n");
			printf("1 1\n");
			continue;
		}
		reverse(a+c,a+d+1);//求一段区间的逆序。头文件为string
		for(i=0;i<n-1;i++){
			if(a[i] < a[i+1]){
				flag3 = 2;
			}
			else{
				flag3 = 1;
				break;
			}
		}
		if(flag3 == 1){
			printf("no\n");
		}
		else{
			printf("yes\n");
			printf("%d %d\n",c+1,d+1);
		}
	}

	return 0;
}