CodeForces 451B Sort the Array(排序，逆序)

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 ofn 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

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

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].

/*这道题的思路是用a数组保存原数组数据，用b数组保存a数组sort之后的数据，从左往右找第一个不满足区间的左端点，从右向左找第一个不满足区间的右端点，在验证a数组的不满足区间是否是一个递减的区间，如果是递减的区间则表明可以进行段逆转使整个a是递增的。我们要注意原序列存在原本就是有序的情况，这个时候left会大于n，输出yes和1,1就可以了*/#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const int maxn=100000+10;int a[maxn],b[maxn];int n;int main(){    int i,left,right;    while(cin>>n)    {        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);            b[i]=a[i];        }        sort(b+1,b+n+1);        for(left=1;left<=n;left++)        {            if(a[left]!=b[left]){break;}        }        /*如果left>n则表明原序列a已经是一个递增序列，否则不是递增序列*/        if(left>n){cout<<"yes"<<endl;cout<<1<<" "<<1<<endl;}        else            {/*如果原序列a不是递增的，那么我们还要从右边开始寻找right*/                for(right=n;right>0;right--)                {                    if(a[right]!=b[right]){break;}                }                /*在找到了left和right之后我们要看看a[left]~a[right]区间的元素是不是递减，                如果是递减的输出yes，证明可以转换为一个递增的序列，输出yes和left，right                否则输出no*/                for(i=left;i<right;i++)                {                    if(a[i]<=a[i+1]){cout<<"no"<<endl;break;}                }                if(i==right){cout<<"yes"<<endl;cout<<left<<" "<<right<<endl;}            }    }    return 0;}