ACM 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): 45217 Accepted Submission(s): 16055

 

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

 

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JGShining

 

#include<stdio.h>
#include<string.h>
int main()
{
    char co[1001][16];
    int i, num, max, j, sum[1001];
    char temp[16];
    while(1)
    {
        memset(sum, 0, sizeof(sum));
        max = 0;
        scanf("%d", &num);
        if(num)
        {
            for(i=0; i < num; ++i)
            {
                scanf("%s", temp);
                for(j=0; j<i; j++)
                    if(strcmp(co[j], temp) == 0)break;
                if(j >= i)
                    strcpy(co[j], temp);
                sum[j]++;
                if(sum[max] < sum[j]) max = j;
            }
            printf("%s\n", co[max]);
        }
        else break;
     }
     return 0;
}

 

 

 

解题情况：

TEL的情况是因为在scanf后加上了fflush(stdin)，这个问题还搞不懂，只是知道C和C++标准中从无定义过fflush；

WA的情况是因为将memset函数放在了第一for循环后面，忽略了编译器自动将整型数组默认为0。