Codeforces Round #258 (Div. 2) . Sort the Array 贪心

B. Sort the Array

题目连接：

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

Description

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 Input

3

3 2 1

Sample Output

yes

1 3

Hint

题意

给你n个数，然后你可以交换一个子串的顺序，问你能不能使得这n个数是单调递增的。

题解：

贪心去交换就好了，交换完再check一下就行。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
int n;
int a[maxn];
bool check()
{
    for(int i=2;i<=n;i++)
        if(a[i]<a[i-1])return false;
    return true;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
    {
        int j=i+1;
        for(;j<=n;j++)
            if(a[j]>=a[i])
                break;
        if(j!=i+1)
        {
            reverse(a+i,a+j);
            if(check())
            {
                printf("yes\n");
                printf("%d %d",i,j-1);
            }
            else
                printf("no\n");
            return 0;
        }
    }
    if(check())
    {
        printf("yes\n");
        printf("%d %d",1,1);
    }
    else
        printf("no\n");
}