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  • CodeForces 451B

    Sort the Array

    Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

    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

    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

     1 //2016.8.2
     2 #include<iostream>
     3 #include<cstdio>
     4 #include<algorithm>
     5 
     6 using namespace std;
     7 
     8 int main()
     9 {
    10     long long arr_src[100005], arr_dest[100005];
    11     int l, r, n;
    12     while(cin >> n)
    13     {
    14         r = l = 1;
    15         for(int i = 1; i <= n; i++)
    16         {
    17             scanf("%lld", &arr_src[i]);
    18             arr_dest[i] = arr_src[i];
    19         }
    20         sort(arr_dest+1, arr_dest+1+n);
    21         for(int i = 1; i <= n; i++)
    22           if(arr_src[i] != arr_dest[i])
    23           {
    24               l = i;
    25               break;
    26           }
    27         for(int i = n; i >= 1; i--)
    28             if(arr_src[i]!=arr_dest[i])
    29             {
    30                 r = i;
    31                 break;
    32             }
    33         bool fg = true;
    34         reverse(arr_dest+l, arr_dest+1+r);
    35         for(int i = 1; i <= n; i++)
    36         {
    37             if(arr_src[i] != arr_dest[i])
    38             {
    39                 fg = false;
    40                 break;
    41             }
    42         }
    43         if(fg)printf("yes
    %d %d
    ", l, r);
    44         else printf("no
    ");
    45     }
    46     return 0;
    47 }
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  • 原文地址:https://www.cnblogs.com/Penn000/p/5758292.html
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