logic is very simple ...just trace the program you will understand.
On Wed, Jul 25, 2012 at 7:28 PM, Navin Kumar <algorithm.i...@gmail.com>wrote:
> #include <stdio.h>
> #include <stdlib.h>
> #include <ctype.h>
>
> char *decrypt(char *s)
> {
> int n;
> char *digit = (char *)malloc(sizeof(char) * 5);
> char *p1 = (char *)malloc(sizeof(char) * 1024);
> char *p = p1;
>
> char *digit2;
> char prev;
> prev = *s;
>
> *p++ = *s++;
>
> while(s != '\0') {
> if(isdigit(*s)) {
> digit2 = digit;
> while(isdigit(*s))
> *digit2++ = *s++;
> *digit2 = '\0';
> n = atoi(digit);
> while(n > 1) {
> *p++ = prev;
> n--;
>
> }
> }
> prev = *s;
> if(*s != '\0')
> *p++ = *s++;
> else{
> *p = *s;
> break;
> }
>
>
> }
> *p = '\0';
> return p1;
> }
>
> int main()
> {
> char *s = "a10b10c1d3e4";
> char *s2 = decrypt(s);
>
> printf("%s", s2);
> return 0;
>
> }
>
> On Wed, Jul 25, 2012 at 6:46 AM, Sathish babu <satbrucei...@gmail.com>wrote:
>
>> can anyone provide the code to convert ab1cd3 to abcddd
>> **~Sathish Babu~**
>>
>>
>>
>> On Tue, Jul 24, 2012 at 11:39 PM, Mind Boggler <min.b...@gmail.com>wrote:
>>
>>> Traditional decryption problem
>>> Convert a3b2c5 into aaabbccccc
>>> Can anyone Put forward an algo for the test case: a3b1c3d1
>>> Thanx
>>>
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>
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