• hdu1004


    Let the Balloon Rise

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 50088    Accepted Submission(s): 17947

    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
     
     
     
     
     
     
     
     
    1. #include<iostream>
    2. #include<cstdio>
    3. #include<string.h>
    4. using namespace std;
    5. int main()
    6. {
    7.  int sum[1001],i,n,j;
    8.  char co[1001][16];
    9.     char temp[16];
    10.  while(1)
    11.  {
    12.   memset(sum,0,sizeof(sum));
    13.   int max=0;
    14.   scanf("%d",&n);
    15.   if(n)
    16.   {
    17.   for(i=0;i<n;i++)
    18.   {
    19.    scanf("%s",temp);
    20.    for(j=0;j<i;j++)
    21.    {
    22.     if(strcmp(co[j],temp)==0)
    23.      break;
    24.    }
    25.    if(j>=i)
    26.     strcpy(co[j],temp);
    27.    sum[j]++;
    28.    if(sum[max]<sum[j])
    29.     max=j;
    30.   }
    31.   printf("%s\n",co[max]);
    32.   }
    33.   else
    34.    break;
    35.  }
    36.  return 0;
    37. }
  • 相关阅读:
    关于mysql 中schema的相关操作
    ASP.NET Core MVC 之过滤器(Filter)
    docker-Error: No such container
    java多种加密和解密方式
    linux 下的rpm 和ivh各是什么意思
    新生赛题目准备
    Figma 学习笔记 – Keyboard Shortcut and Tips 小技巧
    Figma 学习笔记 – Frame
    Chart.js 学习笔记
    SQL Server STRING_AGG
  • 原文地址:https://www.cnblogs.com/Deng1185246160/p/2941091.html
Copyright © 2020-2023  润新知