zoukankan      html  css  js  c++  java
  • A1029. Median

    Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median of S2={9, 10, 15, 16, 17} is 15. The median of two sequences is defined to be the median of the nondecreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.

    Given two increasing sequences of integers, you are asked to find their median.

    Input

    Each input file contains one test case. Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer N (<=1000000) is the size of that sequence. Then N integers follow, separated by a space. It is guaranteed that all the integers are in the range of long int.

    Output

    For each test case you should output the median of the two given sequences in a line.

    Sample Input

    4 11 12 13 14
    5 9 10 15 16 17
    

    Sample Output

    13

     1 #include<cstdio>
     2 #include<iostream>
     3 #include<algorithm>
     4 using namespace std;
     5 int num1[1000001], num2[1000001];
     6 int main(){
     7     int N1, N2;
     8     scanf("%d", &N1);
     9     for(int i = 0; i < N1; i++){
    10         scanf("%d", &num1[i]);
    11     }
    12     scanf("%d", &N2);
    13     for(int i = 0; i < N2; i++){
    14         scanf("%d", &num2[i]);
    15     }
    16     int index = -1, ans = 0, i = 0, j = 0, mid = (N1 + N2 - 1) / 2;
    17     while(i < N1 && j < N2 && index != mid){
    18         if(num1[i] < num2[j])
    19             ans = num1[i++];
    20         else ans = num2[j++];
    21         index++;
    22     }
    23     while(i < N1 && index != mid){
    24         ans = num1[i++];
    25         index++;
    26     }
    27     while(j < N2 && index != mid){
    28         ans = num2[j++];
    29         index++;
    30     }
    31     printf("%d", ans);
    32     cin >> N1;
    33     return 0;
    34 }
    View Code
  • 相关阅读:
    Guru's Guide to SQL Server Architecture and Internals
    如何诊断修正17883,17884,17887,17888错误
    Debugging a SQL Server query with WinDbg
    Windbg调试Sql Server 进程
    SQL SERVER 技术博客 外文
    sql server book
    SQL Server Debugging with WinDbg – an Introduction
    Concurrency(Locking, Blocking and Row Versioning)
    如何自学Android, 教大家玩爆Android
    Storage protocol stacks
  • 原文地址:https://www.cnblogs.com/zhuqiwei-blog/p/8513944.html
Copyright © 2011-2022 走看看