ACM训练二B题

这是比赛后打的题目，思路很清晰：申明一个结构体，将输入的数复制在这个结构体数组中，排序后比对下标，找到变动的首下标和尾下标，再看这段是否逆序了。

Being a programmer, you like arrays a lot. For your birthday, your friends have given you an array a consisting of ndistinct 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 Input

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

#include <iostream>
#include<algorithm>
using namespace std;

int a[100017];

struct NUM{
 int x,id;
}b[100017];

bool cmp(NUM x,NUM y)
{

    return x.x < y.x;
}

int main()
{
    int n;
    cin >> n;
    for(int i = 1;i <= n;i++)
    {

        cin >> a[i];
        b[i].x = a[i];
        b[i].id = i;
    }

    sort(b+1,b+n+1,cmp);

    int i,j;
    for(i = 1;i <= n;i++)
    {

        if(b[i].id != i)
        {
            break;
        }
    }

    if(i == n+1)
    {

        cout << "yes" << endl << "1  1" << endl;
        return 0;
    }
    for(j = n;j > 0;j--)
    {

        if(b[j].id != j)
        {

            break;
        }
    }

    int tt = i;int t = j;
    while(i<=j)
    {

        if(a[i] != b[t--].x)
        {

            break;

        }
        i++;
    }

    if(i == j+1)
    {

        cout << "yes" << endl << tt << "  " << j << endl;
    }
    else
    {

        cout << "no" << endl;
    }
    return 0;
}