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  • Codeforces Round #258 (Div. 2/B)/Codeforces451B_Sort the Array

    解题报告

    http://blog.csdn.net/juncoder/article/details/38102391

    对于给定的数组,取对数组中的一段进行翻转,问翻转后是否是递增有序的。

    思路:

    仅仅要找到最初递减的区域,记录区域内最大和最小的值,和区间位置。

    然后把最大值与区间的下一个元素对照,最小值与区间上一个元素对照。

    这样还不够,可能会出现两个或两个以上的递减区间,这样的情况直接pass,由于仅仅能翻转一次。

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #define inf 999999999999
    #define LL long long
    #define swap(x,y,t) ((t)=(x),(x)=(y),(y)=(t))
    using namespace std;
    LL num[100010];
    int main()
    {
        int n,i,j;
        while(cin>>n)
        {
            for(i=1; i<=n; i++)
                cin>>num[i];
            for(i=1; i<=n; i++)
            {
                if(num[i+1]<num[i])
                    break;
            }
            num[n+1]=inf;
            if(i==n+1)
            {
                printf("1 1
    ");
                continue;
            }
            int l,r;
            LL maxx=0,minn=inf;
            int t=i;
            for(; i<=n; i++)
            {
                if(maxx<num[i])
                {
                    maxx=num[i];
                    r=i;
                }
                if(minn>num[i])
                {
                    l=i;
                    minn=num[i];
                }
                if(num[i+1]>num[i])
                    break;
            }
            int u=i;
            int q=0;
            for(; i<=n; i++)
            {
                if(num[i+1]<num[i])
                {
                    q=1;
                    printf("no
    ");
                    break;
                }
            }
            if(!q)
                if(num[u+1]>=maxx&&num[t-1]<minn)
                {
                    printf("yes
    ");
                    if(l<r)
                        swap(l,r,q);
                    printf("%d %d
    ",r,l);
                }
                else printf("no
    ");
        }
        return 0;
    }
    

     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].


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  • 原文地址:https://www.cnblogs.com/bhlsheji/p/4198887.html
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