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].

题意：给你一个序列，问翻转一个区间的数，即将那个区间的数倒过来后，整个序列是否是单调递增的

思路：直接扫一遍寻找递减的那个区间，然后排序检查就好了

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
int a[MAXN];
int main()
{
int n,i;
while(scanf("%d",&n)!=EOF)
{
int bz=0,k=-1;
int l=1,r=1;
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
k=a[1];
for(i=2;i<=n;i++)
{
if(a[i]<k)
{
if(bz==0)
l=i-1;
r=i;
bz=1;
k=a[i];
}
else
{
k=a[i];
if(bz)
break;
}
}
//printf("%d %d\n",l,r);
sort(a+l,a+r+1);
for(i=2;i<=n;i++)
if(a[i]<a[i-1])
break;
if(i>n)
printf("yes\n%d %d\n",l,r);
else
printf("no\n");
}
return 0;
}