[解題報告]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): 50554    Accepted Submission(s): 18117

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

ZJCPC2004

 

 

Recommend

JGShining

 

 

 

 

 

 

 

 

 

用color[1000][16]來存儲顏色信息

用num[1000]來統計每個顏色出現的次數

先輸入一個顏色，從第二個顏色的輸入開始，每輸入一個，都要和之前輸入的所有顏色進行比較，若是一樣，則在數組對應位置上+1，

如

3

color[0] pink    

color[1] pink num[1]=2

color[2] blue num[2]=1

然后在num[1000]中查找最大數，輸出其下標，找到對應的顏色輸出

 

#include <stdio.h>
#include <string.h>
int main()
{
    int n,i,j,num[1000];
    int max=0,t=0;
    char color[1000][16];
    while(scanf("%d",&n)!=EOF)
    {
        if(n)
        {
            num[0]=0;
            scanf("%s",color[0]);
            for(i=1;i<n;i++)
            {
                num[i]=0;
                scanf("%s",color[i]);
                for(j=0;j<i-1;j++)
                    if(strcmp(color[i],color[j])==0) num[i]+=1;
            }
            max=0;
            t=0;
            for(i=1;i<n;i++)
               if(max<num[i]) 
               {
                   max=num[i];
                   t=i;}
            printf("%s\n",color[t]);
        }
    }
}