杭電ACM 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): 118927    Accepted Submission(s): 46618

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  

Recommend JGShining   |   We have carefully selected several similar problems for you:   1002  1001  1008  1005  1009   

#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int main(void)
{
	int n, i, j;
	cin >> n;
	while (n)
	{
		int Max = -1000000;
		int maxc = 0;
		string *color = new string[n];
		int * sum = new int[n];
		int * visited = new int[n]; //增加一個監視數組，防止重複計數
		memset(sum, 0, n*sizeof(int));
		memset(visited, 0, n*sizeof(int));
		for (int i = 0; i < n; i++)
		{
			cin >> color[i];
		}
		for (i = 0; i < n; i++)
		{
			if (visited[i] == 1)//已經記錄過就跳過本次
				continue;
			else
			{
				visited[i] = 1;//進行記錄，先指派為1
				sum[i]++;//初始化值為1，因為進行到這一步就意味着已經有一個這樣的顔色了
				for (j = i + 1; j < n; j++)
					if (visited[j] == 0 && !(color[i].compare(color[j]))) //.compare操作相同傳回0，！0=1，并且還要保證沒有被浏覽過
					{
						sum[i]++;
						visited[j] = 1;//标記此顔色已被記錄過
					}
			}
		}
		for (int k = 0; k < n; k++)
			if (sum[k] > Max) {
				Max = sum[k];
				maxc = k;
			}//記錄最大值的坐标
		cout <<color[maxc] << endl;  //将坐标直接帶入到string color的坐标中（因為sum和color的坐标是相對的）
		cin >> n;
	}
}
           

簡單的周遊，處理好一些小細節就可以