杭电oj 1004------Let the Balloon Rise

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Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 153132    Accepted Submission(s): 60752

 

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

 用二维数组存储颜色

#include <stdio.h>
#include <string.h>
int main()
{
    int n,i,j,max,t,a[1100];
    char s[1100][20];
    while(~scanf("%d",&n),n)
    {
        max=0;
        t=0;
        memset(a,0,sizeof(a));      //初始化数组a
        for(i=0;i<n;i++)
        { 
            scanf("%s",s[i]);       //输入颜色
        }
        getchar();                  //清除回车
        for(i=0;i<n;i++)
        {
            for(j=i+1;j<n;j++)
            {
                if(strcmp(s[i],s[j])==0)    //逐次比较
                    a[i]++;                 //一样a[i]++
            }                               //以样例 5
            if(max<a[i])                    //       green
            {                               //       red
                max=a[i];                   //       blue
                t=i;      //最多次下标       //       red
            }                               //       red       这时候 a[0]~a[4]的数分别为0 2 0 1 0 
                                            // 2的下标为1  就是red出现最多次
        }
        printf("%s",s[t]);
        printf("\n");
    }
    return 0;
}