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Encoding

2024-11-01 09:55:39   205人阅读

Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28533 Accepted Submission(s): 12642


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
源代码一：考虑字符串数组中的字符结尾
#include <stdio.h>#include <string.h>#include <stdlib.h>char stra[10000];int main(){  int i,j,n,lena,count;  scanf("%d",&n);  getchar();  while(n--)  {     scanf("%s",stra);    count=1;    lena=strlen(stra);    for(i=0,j=0;i<lena-1;i++)      if(stra[i]==stra[i+1])        count++;       else      {        if(count==1)          printf("%c",stra[i]);          else         {          printf("%d%c",count,stra[i]);            count=1;         }         }    if((i==lena-1) && count>1)       printf("%d",count);         printf("%c\n",stra[i]);        }  system("pause");  return 0;    }
源代码二：直接使用字符数组
#include <stdio.h>#include <string.h>#include <stdlib.h>char a[10001];int main(){  int n,len,i,count;  scanf("%d",&n);  getchar();  while(n--)  {    count=1;    gets(a);    len=strlen(a);    for(i=0;i<len;i++)    {      if(a[i]==a[i+1])        count++;      else      {	if(count==1)	  printf("%c",a[i]);	else	{	  printf("%d%c",count,a[i]);	  count=1;	}      }
    }    printf("\n");   }   system("pause");   return 0;}

Encoding


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