题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1004
Let the Balloon Rise
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 90295 Accepted Submission(s): 34294
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.
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.
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.
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
green
red
blue
red
red
3
pink
orange
pink
0
Sample Output
red
pink
pink
Author
WU, Jiazhi
Source
联系map的用法
#include<cstdio>
#include<cstring>
#include<string>
#include<map>
#include<iostream>
using namespace std;
#define N 1010
map <string,int> mp;
struct MAX {
int a;
string color;
};
int main()
{
int n;
while(~scanf("%d",&n),n)
{
mp.clear();
for(int i = ;i < n ;i++)
{
char ss[];
scanf("%s",ss);
string s = ss;
if(mp.find(ss)==mp.end()) mp[s] = ;
else mp[s]++;
}
MAX mx;
int sum = ;
map <string , int >:: iterator it;
for( it = mp.begin(); it!=mp.end(); it++)
{
if(sum<(*it).second)
{
sum = (*it).second;
mx.a = sum;
mx.color = (*it).first;
}
}
printf("%s\n",mx.color.c_str());
}
return ;
}