HDU-1020

Encoding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40125    Accepted Submission(s): 17801

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

 

0ms

#include <stdio.h>
#include<string.h>
#define M 10005
char a[M];
char b[M];
char c[M];
int main()
{
    int n;
    int i, j, k = 0, p = 0, q, z;
    scanf("%d", &n);
    while (n--)
    {   
   scanf("%s",a);
        p = 0;
        for (i = 0; i<strlen(a);i=j)
        {
            j = i + 1;
            k = 1;
            while (a[i] == a[j]) {j++;k++;}
                if (k != 1)
                {    q = 0;
                    while (k)
                    {
                        c[q++] = k % 10 + '0';
                        k = k / 10;
                    }
                    for (z = q - 1; z >= 0; z--)
                        b[p++] = c[z];
                    b[p++] = a[i];

                }
                else b[p++] = a[i];
        }
        b[p] = '\0';
        printf("%s\n", b);
    }
    return 0;
}

#include <stdio.h>
#include<string.h>
#define M 10005
char a[M];
int main()
{
 int n;
 int i,j,k;
 scanf("%d", &n);
 while (n--)
 {
  int num=1;
  scanf("%s",a);
  for (i = 0; i<strlen(a);i++)
  {
      while(a[i]==a[i+1]){ num++; i++; }
   if(num==1) printf("%c",a[i]);
   else {printf("%d%c",num,a[i]);num=1;}  
  }
  printf("\n");
 
 }
 return 0;
}