OJ Board

2 10 10101
5 3 126
15 11 A4C
16 10 FFFFFFFFFF
16 2 FFFFFFFFFF
2 10 0
10 3 1234567
2 15 1001102001
15 2 3014593F19
10 10 1234567890
16 16 FEDCBA0987
2 3 0
2 10 0
10 2 0
14 8 0

Output i got was :
10101 base 2 = 21 base 10
126 is an illegal base 5 number
A4C base 15 = 1821 base 11 ...

Sorry, Sillly mistake, i got AC

Hello,
This is my first ACM prog.
This is what I took care:-
1. Long Integers (used unsigned long long).
2. Took care of Minimum base of a number.
3. Took help again from some of others I knew , in vain though.
Below is the code:-

#include <stdio.h>
#include <string.h>
#include <stdlib.h ...