杭电ACM-HDU1004-Let the Balloon Rise

题目来自杭电ACM: acm.hdu.edu.cn

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 93510    Accepted Submission(s): 35712

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

5greenredblueredred3pinkorangepink0

 

Sample Output

redpink

 

Author

WU, Jiazhi

 

Source

ZJCPC2004

代码如下：

#include <stdio.h>#include <string.h>#define N 1010int count[N];char ballon[N][16];int main(){int n, i, j;while (1){scanf("%d", &n);if (n == 0){break;}gets(ballon[0]);for (i = 1; i <= n; ++i){gets(ballon[i]);}int max = 1, position = 1;for (i = 1; i <= n; ++i){count[i] = 0;for (j = i; j <= n; ++j){if (strcmp(ballon[i], ballon[j]) == 0){count[i]++;if (count[i] > max){max = count[i];position = i;}}}}printf("%s\n", ballon[position]);}return 0;}

这是一个较为简单的考察字符串的题目，通过头文件string.h中的strcmp函数来判断字符串是否相同。