acm集训训练赛B题【排序+模拟】

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.

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).

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 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].

二 、题目源程序（一）

#include<bits/stdc++.h>
#define maxn 100005
typedef long long LL;
using namespace std;
int a[maxn];
int main()
{
    int n,x=1,y=1;
    cin>>n;
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=1;i<n;i++){
        if(a[i]>a[i+1]){
            x=i;break;
        }
    }
    for(int i=n;i>=1;i--){
        if(a[i]<a[i-1]){
            y=i;
            break;
        }
    }
    for(int i=0;i<y-x;i++)
    {
        int t;
        t=a[x+i];
        a[x+i]=a[y-i];
        a[y-i]=t;
    }
    int flag=0;
    for(int i=1;i<n;i++){
        if(a[i]>a[i+1]){
            flag=1;
            break;
        }
    }
    if(flag) cout<<"no"<<endl;
    else{
        cout<<"yes"<<endl<<x<<" "<<y<<endl;
    }
    return 0;
}

题目源程序（二）

#include<bits/stdc++.h>
#define maxn 100005
typedef long long LL;
using namespace std;
int a[maxn];
int main()
{
    int n,x=1,y=1;
    cin>>n;
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=1;i<n;i++){
        if(a[i]>a[i+1]){
            x=i;break;
        }
    }
    for(int i=n;i>=1;i--){
        if(a[i]<a[i-1]){
            y=i;
            break;
        }
    }
    reverse(a+x,a+y+1);
    int flag=0;
    for(int i=1;i<n;i++){
        if(a[i]>a[i+1]){
            flag=1;
            break;
        }
    }
    if(flag) cout<<"no"<<endl;
    else{
        cout<<"yes"<<endl<<x<<" "<<y<<endl;
    }
    return 0;
}

三、解题思路

题意分析：

给定一个序列，能否通过逆序它的一个子序列使得整个序列单调递增，如果能，输出需要逆序的下标，否则输出no

思路：

1、从左到右遍历，第一次出现比前一个数小的下标；

2、然后从右往左遍历，第一次出现比后一个数小的下标；

3、将这段区间逆序，然后判断整个序列是否单调递增。

四、心得体会

（1）reverse函数的运用，包含于头文件<algorithm>中。

（2）这是一道比较简单的排序与模拟题。