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

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最新推荐文章于 2025-08-01 11:12:48 发布

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本文介绍了一种通过翻转数组中的某一段来实现排序的方法。该算法首先判断数组是否可以通过一次翻转操作变为有序，并输出翻转的起始和结束位置。文章提供了完整的算法实现代码及示例。

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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].

醉了，写了半天不对，原来又是把题目看错了。。翻转序列中某一段能否使得序列递增

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string>
#include <cstring>
#include <cmath>
#define ll __int64
ll a[100009];
ll b[100009];

using namespace std;

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof a);

        for(int i=1;i<=n;i++)
        {
            scanf("%I64d",&a[i]);
            b[i]=a[i];
        }

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

        int st=1,en=1;

        for(int i=1;i<=n;i++)
        {
            if(a[i]!=b[i])
            {
                st=i;
                break;
            }
        }

        for(int i=n;i>=1;i--)
        {
            if(a[i]!=b[i])
            {
                en=i;
                break;
            }
        }

        for(int i=0;i<(en-st+1)/2;i++)
        {
            swap(a[en-i],a[st+i]);
        }

        int f=0;

        for(int i=1;i<=en;i++)
        {
            if(a[i]!=b[i])
            {
                f=1;
                break;
            }
        }

        if(f)
        cout<<"no"<<endl;
        else
        {
            cout<<"yes"<<endl;
            cout<<st<<" "<<en<<endl;
        }

    }
    return 0;
}