hdoj 1020 Encoding

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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

2ABCABBCCC

Sample Output

ABCA2B3C

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;}

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