CodeForces 451B Sort the Array

J - Sort the Array

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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

33 2 1

Output

yes1 3

Input

42 1 3 4

Output

yes1 2

Input

43 1 2 4

Output

no

Input

21 2

Output

yes1 1

Hint

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：判断能否找到递减的子列，将其翻转后得到的整个数列递增，只能找一次，最后输出，如果能还要输出翻转的首尾位

#include<stdio.h>int n,a[100005];int main(){    int i,j,h;    bool bo=true;    scanf("%d",&n);    for (i=0;i<n;i++)scanf("%d",&a[i]);    i=0;  for(i=0;i<n-1;i++)//找到开始递减的数a[i],即左端点{if(a[i]>a[i+1]){            break;}}if (i==n-1)      //如果数列一直递增，就省事多了    {        bo=false;        printf("yes\n1 1");    }      for(j=i;j<n-1;j++)//找到递减序列的右端点{if(a[j]<a[j+1]){break;}}    if (i>0&&a[j]<a[i-1]&&bo) //看看能不能接上，就是递减子列//翻转后的左端点不能小于前一个数    {        printf("no\n");        bo=false;    }    if (j==n-1&&bo)       //判断后面是否还有数    {        bo=false;        printf("yes\n%d %d\n",i+1,j+1);    }    h=j;    if (h<n-1&&a[h+1]<a[i]&&bo)  //看看能不能接上    {            printf("no\n");            bo=false;    }    while(h<n-1&&bo)    //如果后面还有数有，看看能不能保证递增    {        if (a[h]>a[h+1])        {            bo=false;            printf("no\n");        }        h++;    }    if (bo) printf("yes\n%d %d\n",i+1,j+1);      return 0;}