Let the Balloon Rise http://acm.hdu.edu.cn/showproblem.php?pid=1004

本文提供了一个关于LettheBalloonRise竞赛题目的解决方案,通过使用C语言实现,该程序能够处理多组测试数据,统计并输出出现次数最多的气球颜色。解答过程中涉及字符串比较和复制等关键操作。

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58098    Accepted Submission(s): 21290


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
 

 

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#include<stdio.h>
#include<string.h>
int main()
{
	int i,n;
	char a[1000][20];
	char b[20];
	while(scanf("%d",&n)&&n!=0)
	{
		int j,k,q,p;
		int max=0;
		for(i=0;i<n;i++)
		{
			scanf("%s",a[i]);
		}
		for(i=0;i<n;i++)
		{
			k=0;
			for(j=i;j<n;j++)
			{
				p=strcmp(a[i],a[j]);		
				if(p==0)
					k++;
			}
			if(max<k)
			{
				max=k;
				strcpy(b,a[i]);
			}			
		}
		printf("%s\n",b);	
	}
	return 0;
}


编这段代码时有个有趣的事,就是这段代码if(max<k){max=k;strcpy(b,a[i]);} }printf("%s\n",b); 第一次我是这么编的但是一直就出现烫,后来啊,我重新写了了一遍什么事也没了。解决此题的关键是strcpy函数和strcmp函数。再感到难的是循环及判断条件。

转载于:https://www.cnblogs.com/wangyouxuan/p/3239369.html

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