Let the Balloon Rise 1004

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 57360    Accepted Submission(s): 20952

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

5greenredblueredred3pinkorangepink0

 

Sample Output

redpink

 

Author

WU, Jiazhi

 

Source

ZJCPC2004

 

Recommend

JGShining

用容器map一下就解决了

#include<stdio.h>
#include<cstring>
#include<iostream>
#include<map>
using namespace std;
const int MAXN=1000;
int main()
{
    map<string,int>color;
    string cnt;
    string index;
    int i;
    int n;
    int max;
    while(scanf("%d",&n),n)
    {    
       max=0;
        while(n--)
        {
        
            cin>>cnt;
            color[cnt]++;
            if(color[cnt]>max)
            {
            index=cnt;
            max=color[cnt];
            }
            
        }
      cout<<index<<endl;
    }
    return 0;   
}

不用容器的代码：

#include<iostream>
#include<cstring>
using namespace std;
struct node
{
    char s[16];
    int x;
}a1[1000],a2[1000];
int main()
{
    int n,i,j;
    while(~scanf("%d",&n))
    {
        if(n==0) break;
        int k=0,m=0;
        for(i=0;i<n;i++)
        {
            scanf("%s",a1[i].s);
            a1[i].x=1;    
            strcpy(a2[i].s,a1[i].s);
        }
        for(i=0;i<n;i++)
        {
            for(j=i+1;j<n;j++)
            {
                if(!strcmp(a1[i].s,a2[j].s))
                a1[i].x++;
            }
        }
            for(i=0;i<n;i++)
            {
                if(a1[i].x>m){
                    m=a1[i].x;
                    k=i;
                 }
             }
            printf("%s\n",a1[k].s);
    }
    return 0;
}