hdu 1004 Let the Balloon Rise

最新推荐文章于 2021-06-24 22:53:24 发布

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本文介绍了一种解决气球飘浮竞赛中，根据气球颜色统计最流行问题的方法。通过输入气球数量及每种颜色，程序会输出出现频率最高的颜色。示例代码展示了如何实现这一功能。

1、题目描述：

A - Let the Balloon Rise

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 2104

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

2、ac代码：

#include<stdio.h>
#include<string.h>
char a[1005][20];
int count[1005];
int main()
{
    int n,j;
    char str[20];
    while(scanf("%d",&n)!=EOF)
    {

        if(n==0)
            break;
        int m=0;
        getchar();
        memset(count,0,sizeof(count));
        for(int i=0; i<n; i++)
        {
            scanf("%s",str);
            for(j=0; j<m; j++)
            {
                //printf("**");
                if(strcmp(str,a[j])==0)
                {
                    count[j]++;
                    break;
                }
            }
            if(j==m)
            {
                strcpy(a[m],str);
                m++;
            }
        }
        int maxx=-1,cnt;
        for(int i=0; i<m; i++)
        {
            if(count[i]>maxx)
            {
                maxx=count[i];
                cnt=i;
            }
        }
        printf("%s\n",a[cnt]);
    }
    return 0;
}