杭电1004（简单题）

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

此题比较简单，我的做法是先对字符串排序，排序后相同的字符串必定紧挨着一起。要计算出每个字符出现的次数，只要计算其下面有多少个字符跟它相同，然后再加一即可。

#include<stdio.h>#include<string.h>#include<stdlib.h>char ballon[1010][20];int cmp(const void *_a,const void *_b){    char *a=(char *)_a;    char *b=(char *)_b;    return strcmp(a,b);}int main(){    int N,max,mindex,t,tindex;//max是字符串出现的次数的最大值，mindex是次数最多的字符串的下标。t是当前扫描的字符串出现的次数，tindex是与当前字符串相同，且第一次出现的字符串的下标    while(scanf("%d",&N)==1&&N>0)    {        max=0;        mindex=0;        int i,j;        for(i=0;i<N;i++)        {            memset(ballon[i],0,sizeof(char)*20);            scanf("%s",ballon[i]);        }        qsort(ballon,N,sizeof(char)*20,cmp);        for(i=0;i<N;i++)        {            t=1;            tindex=i;            for(j=i+1;;j++)            {                if(strcmp(ballon[tindex],ballon[j])!=0)                    break;                t++;            }            i=j-1;//起初i=j，一直WA。因为j就是下一次开始计数的起点。如果i=j,下一轮循环i++后，i=i+1.那么下一轮计数就会少计数一个            if(t>max)            {                max=t;                mindex=tindex;            }        }        printf("%s\n",ballon[mindex]);    }    return 0;}