hdoj 1020 Encoding

http://acm.hdu.edu.cn/showproblem.php?pid=1020

Encoding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26061    Accepted Submission(s): 11465

Problem Description

Given a string containing only 'A' - 'Z', we could encode it using the following method:

1. Each sub-string containing k same characters should be encoded to "kX" where "X" is the only character in this sub-string.

2. If the length of the sub-string is 1, '1' should be ignored.

 

 

Input

The first line contains an integer N (1 <= N <= 100) which indicates the number of test cases. The next N lines contain N strings. Each string consists of only 'A' - 'Z' and the length is less than 10000.

 

 

Output

For each test case, output the encoded string in a line.

 

 

Sample Input

  
  

   
   2
ABC
ABBCCC
  
  

 

 

Sample Output

  
  

   
   ABC
A2B3C
  
  

 

第一次写的，理解错误，想复杂了，输入是是AABBCCAAB，输出应该是2A2B2C2AB,理解成了输出4A3B2C

#include<stdio.h>
#include<string.h>
int main()
{
    int N,n,i,j,k,b[30],c[30];
    char s[200000],a[]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
    scanf("%d",&N);
    while(N--)
    {
          scanf("%s",s);    
          n=strlen(s);
          memset(b,0,sizeof(b));          
          for(i=0,k=0;i<n;i++)
          {
                for(j=0;j<26;j++)
                {
                    if(s[i]==a[j])
                    {
                         b[j]+=1;
                         break;
                    }
                }
                c[k]=j;
                if(b[j]<=1)
                k+=1;
          }
          //printf("%d\n",k);
          
          for(i=0;i<k;i++)
          {
               if(b[c[i]]!=0)
               {
                   if(b[c[i]]==1)
                   printf("%c",a[c[i]]);
                   else
                   printf("%d%c",b[c[i]],a[c[i]]);
               }
          }
          printf("\n");       
    }
    //while(1);
    return 0;
}                                              

这是理解正确的代码

#include<stdio.h>
#include<string.h>
int main()
{
    int N,n,m,i,j,k;
    char s[20000],a[]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
    scanf("%d",&N);
    while(N--)
    {
            scanf("%s",s);
            n=strlen(s);
            for(i=0,m=0;i<n;i++)
            {       
                  for(j=0;j<26;j++)
                  {
                      if(s[i]==a[j])
                      m+=1;
                  }
                  if(s[i]!=s[i+1]&&m!=0)
                  {
                        if(m==1)
                        printf("%c",s[i]);
                        else
                        printf("%d%c",m,s[i]);
                        m=0;
                  }
            }
            printf("\n");
    }
    //while(1);
    return 0;
}