HDOJ1004 Let the Balloon Rise_C++

最新推荐文章于 2021-02-27 19:36:55 发布

Luna990105

最新推荐文章于 2021-02-27 19:36:55 发布

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本文链接：https://blog.youkuaiyun.com/Luna990105/article/details/79270616

本文介绍了一个简单的编程问题：统计竞赛中发放的各种颜色气球的数量，并找出最流行的气球颜色。文章提供了一段C++代码实现，该代码使用数组来记录每种颜色气球的出现次数。

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Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/32768 K (Java/Others)

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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

统计每个字符串出现的次数，找出最大的次数。这里设置两个数组，一个存放颜色信息，一个存放出现次数，两个数组根据下标一一对应。（不好意思又穷举了，有想过动态考虑，应该是可以的，留一个问题在这待解决。）

#include<iostream>

#include<string>

#include<string.h>
using namespace std;
int main()
{

int n;
while(cin>>n&&n!=0)
{
string color[1000];
int num[1000];
for(int i=0;i<n;i++)
{
cin>>color[i];
num[i] =1;
}

for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
if(color[i]==color[j]) num[i]++; //strcmp(color[i],color[j]==0)
}
}
int max=0,maxnum;
for(int i=0;i<n;i++)
{
if(num[i]>max)
{
max=num[i];
maxnum=i;
}
}
cout<<color[maxnum]<<endl;

}

return 0;
}