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【ACM】hdu_1004_Let the Balloon Rise

發(fā)布時(shí)間：2023/10/12 编程问答 114 如意码农
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Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 57556    Accepted Submission(s): 21037

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

hdu_1004_Let the Balloon Rise_201307271045.c

#include <stdio.h>
#include <string.h>
int main()
{
 char str[1010][18];
 int n;
 while(scanf("%d",&n),n)
 {
 int i,j,k,m=0,num=0;
 char str1[18],str2[18];
 for(i=0;i<n;i++)
 {
  gets(str[i]);
 }
 for(j=1;j<n;j++)
 for(i=0;i<n-j;i++)
 if(strcmp(str[i],str[i+1])>0)
 {
  strcpy(str1,str[i]);
  strcpy(str[i],str[i+1]);
  strcpy(str[i+1],str1);
 }
 strcpy(str2,str[0]);
    for(i=0;i<n;i++)
    {
     if(strcmp(str2,str[i])==0)
     num++;
     else
     {
      strcpy(str2,str[i]);
      num=1;
      if(num>m)
      {
         m=num;
        
         k=i-1;
      }
     }
    }
    if(num>m)
    {
     m=num;        
        k=i-1;
    }
    printf("%s\n",str[k]);
 }       
 return 0; 
}

//超時(shí)

//參考代碼如下：

#include <stdio.h>
#include <string.h>
main(){
    int n, i, j, t, max, num[1000];
    char color[1000][16];
    while(scanf("%d", &n) != EOF){
        if(n){
            num[0]=0;
            scanf("%s", color[0]);
            for(i=1; i <n; i++){
                num[i]=0;
                scanf("%s", color[i]);
                for(j=0; j <i-1; j++)
                    if(strcmp(color[i], color[j])==0) num[i] +=1;
            }
            max=num[0];
            t=0;
            for(i=1; i <n; i++)
               if(max <num[i]) {max =num[i]; t=i;}
            printf("%s\n",color[t]);
        }
    }
}

//
 
改后代碼：
#include <stdio.h>
#include <string.h>
int main()
{
 char str[1010][18]; 
 int n;
 while(scanf("%d",&n),n)
 {
  int i,j,k,max;
  int num[1010]={0};
  k=max=0;
  for(i=0;i<n;i++)
  {
   scanf("%s",str[i]);
   for(j=0;j<i;j++)
   if(strcmp(str[i],str[j])==0)
   num[i]++;   
   if(num[i]>max)
   {
    max=num[i];
    k=i;
   }
  }
  printf("%s\n",str[k]);
 }
 return 0;
}

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