zoukankan      html  css  js  c++  java
  • HDU1029(KB12-B)

    Ignatius and the Princess IV

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)
    Total Submission(s): 30040    Accepted Submission(s): 12812


    Problem Description

    "OK, you are not too bad, em... But you can never pass the next test." feng5166 says.

    "I will tell you an odd number N, and then N integers. There will be a special integer among them, you have to tell me which integer is the special one after I tell you all the integers." feng5166 says.

    "But what is the characteristic of the special integer?" Ignatius asks.

    "The integer will appear at least (N+1)/2 times. If you can't find the right integer, I will kill the Princess, and you will be my dinner, too. Hahahaha....." feng5166 says.

    Can you find the special integer for Ignatius?
     

    Input

    The input contains several test cases. Each test case contains two lines. The first line consists of an odd integer N(1<=N<=999999) which indicate the number of the integers feng5166 will tell our hero. The second line contains the N integers. The input is terminated by the end of file.
     

    Output

    For each test case, you have to output only one line which contains the special number you have found.
     

    Sample Input

    5 1 3 2 3 3 11 1 1 1 1 1 5 5 5 5 5 5 7 1 1 1 1 1 1 1
     

    Sample Output

    3 5 1
     

    Author

    Ignatius.L
     
    排序输出第n/2个。
     1 //2017-03-14
     2 #include <iostream>
     3 #include <cstdio>
     4 #include <cstring>
     5 #include <algorithm>
     6 
     7 using namespace std;
     8 
     9 const int N = 1e6;
    10 int arr[N];
    11 
    12 int main()
    13 {
    14     int n;
    15     while(scanf("%d", &n)!=EOF)
    16     {
    17         for(int i = 0; i < n; i++)
    18               scanf("%d", &arr[i]);
    19         sort(arr, arr+n);
    20         printf("%d
    ", arr[n/2]);
    21     }
    22 
    23     return 0;
    24 }
  • 相关阅读:
    软件设计中的分层模式, 三层开发遵循的原则,分层开发的特点和优势
    什么是jsp?
    在Servlet中如何如何获取请求的参数?
    Servlet的加载(执行过程,原理)和生命周期
    servlet的注册
    什么是servlet容器
    什么是Servlet
    如何访问动态页面——URL
    什么是C/S? Client/server的简写,这里Server指的是DBServer。
    MVC(Model-View-Controller)软件设计模式
  • 原文地址:https://www.cnblogs.com/Penn000/p/6550238.html
Copyright © 2011-2022 走看看