Let the Balloon Rise

题目描述
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.
输入
nput 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.
输出
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.
样例输入
5
green
red
blue
red
red
3
pink
orange
pink
0

样例输出
red
pink

#include<iostream>
#include<string.h>
using namespace std;

typedef struct
{
	char color[15];
	int n;
}Balloon,*Case;

int Search_Balloon(Case &B,int N)
{
	int i,j,k;
	int max,flag;
	char t[N];
	for(i=1;i<N-1;i++)
	{
		strcpy(t,B[i].color);
		for(k=i;k<N-1;k++)
		{
			flag=1;
			for(j=0;j<15&&flag;j++)
			{
				if(t[j]!=B[k+1].color[j])
				{
					flag=0;
				}
				B[i].n=B[i].n+1;
			}
		}
	}
	i=2;
	max=1;
	while(i<N)
	{
		if(B[max].n<B[i].n)
		{
			max=i;
		}
		i++;
	}
	return max;
}

int main(void)
{
	Case B;
	int i,N=1,max;
	while(N!=0)
	{
		cin>>N;
		B=new Balloon[N+1];
		for(i=0;i<N+1;i++)
		{
			gets(B[i].color);
			B[i].n=0;
		}
		max=Search_Balloon(B,N);
		cout<<B[max].color<<endl;
		delete B;
	}
}