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 input

output

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.

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].

—————————————————————切割线———————————————————————————

题意分析：

给定一个序列，是否能通过逆序它的一个子序列使得整个序列单调递增，假设能。输出须要逆序的下标，否则输出no

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，然后推断整个序列是否单调递增

#include<bits/stdc++.h>
#define maxn 100005
typedef long long LL;
using namespace std;
int a[maxn];
int main()
{
    int n,x=1,y=1;
    cin>>n;
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=1;i<n;i++){
        if(a[i]>a[i+1]){
            x=i;break;
        }
    }
    for(int i=n;i>=1;i--){
        if(a[i]<a[i-1]){
            y=i;
            break;
        }
    }
    reverse(a+x,a+y+1);
    int flag=0;
    for(int i=1;i<n;i++){
        if(a[i]>a[i+1]){
            flag=1;
            break;
        }
    }
    if(flag) cout<<"no"<<endl;
    else{
        cout<<"yes"<<endl<<x<<" "<<y<<endl;
    }
    return 0;
}