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
题目大意
输入几种颜色,输出那种颜色最多。
思路
调用STL库中的map,定义
map<string,int>
表示string到int的一种映射,每输入一种颜色map[string]++,最后用迭代器跑一边map,找到值最大的一个输出就好了。
代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
int main()
{
int n;
while(~scanf("%d",&n)&&n)
{
string str;
string tmp;
map<string,int> q;
map<string,int>::iterator it;
for(int i=1;i<=n;i++)
{
cin>>str;
q[str]++;
}
int ans=-1;
for(it=q.begin();it!=q.end();it++)
{
if(it->second>ans)
{
ans=it->second;
tmp=it->first;
}
}
cout<<tmp<<endl;
}
return 0;
}
本文介绍了一道编程竞赛题目,任务是统计多种颜色中出现次数最多的颜色。使用C++结合STL库中的map实现解决方案,通过迭代遍历找出数量最多的颜色。
976

被折叠的 条评论
为什么被折叠?



