杭电1004 ACM

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 70064    Accepted Submission(s): 26055

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

 

计算哪种颜色出现的次数最多，可以使用结构体，定义

struct color
{
char c[20];
int num;
}a[1005];

数组c为字符串“颜色”，num为他出现的次数，一边输入一边统计，每输入一次，就检查之前是否有相同颜色的出现，若有，加上距离当前最近一个同颜色的num值；

最后计较那个num值最大，输出相应结构体中的数组c，也就是“颜色\n”！

我用C写的代码：

#include<stdio.h>
#include<string.h>
struct color
{
char c[20];
int num;
}a[1005];
int main()
{
int i,j;
long long k,l,n;
scanf("%lld",&n);
while(n!=0)
{
for(i=0;i<n;i++)
{
a[i].num=1;
scanf("%s",a[i].c);
for(j=i-1;j>=0;j--)
{
if(strcmp(a[i].c,a[j].c)==0)
{
a[i].num+=a[j].num;
break;
}
}
}
k=a[0].num;
l=0;
for(i=0;i<n;i++)
{
if(a[i].num>k)
{
l=i;
k=a[i].num;
}
}
printf("%s\n",a[l].c);
scanf("%d",&n);
}
return 0;
}