zoukankan      html  css  js  c++  java
  • HDU 1004 Let the Balloon Rise

    Let the Balloon Rise

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 97957    Accepted Submission(s): 37458


    Problem Description
    Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.

    This year, they decide to leave this lovely job to you.
     
    Input
    Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) -- the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.

    A test case with N = 0 terminates the input and this test case is not to be processed.
     
    Output
    For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.
     
    Sample Input
    5 green red blue red red 3 pink orange pink 0
     
    Sample Output
    red pink
     
    Author
    WU, Jiazhi
     
    Source
     
    Recommend
    JGShining
     
     
    简单使用map容器。
     
     1 #include <iostream>
     2 #include <string>
     3 #include <map>
     4 using namespace std;
     5 
     6 map<string,int> m;
     7 int main()
     8 {
     9     int n;
    10     string s,maxs;
    11     while(cin>>n&&n)
    12     {
    13         for(int i=0;i<n;i++)
    14         {
    15             cin>>s;
    16             if(!m.count(s))
    17                 m[s]=1;
    18             else
    19                 m[s]++;
    20         }
    21 
    22         int cmp=0;
    23         map<string,int>::iterator it;
    24         for(it=m.begin();it!=m.end();it++)
    25         {
    26             if(it->second > cmp)
    27             {
    28                 cmp=it->second;
    29                 maxs=it->first;
    30             }
    31         }
    32         m.clear();
    33         cout<<maxs<<endl;
    34     }
    35 }
  • 相关阅读:
    linux内存的使用与page buffer (转)
    基于linux2.6.38.8内核的SDIO/wifi驱动分析(转)
    RamDisk块设备驱动实例开发讲解一
    Linux加密框架设计与实现(转)
    v4l2子系统学习心得
    一句memset引发的疑案
    linux 信号量之SIGNAL 0(转)
    可重入函数
    从ARM VIVT看linux的cache 处理
    内核抢占与preempt_count
  • 原文地址:https://www.cnblogs.com/cumulonimbus/p/5158757.html
Copyright © 2011-2022 走看看