[Challenge] All your bijective base are belong to us

bitm0de · 09-06-2015, 10:16 PM

Here's a quick solution in C:

Code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

#define MAX_BASE 64

static char M[] = "123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz<=>";

const char *base10_to_b(int b, int n, char *buf)
{
  int i = MAX_BASE - 1;
  while (n)
  {
    --n;
    buf[i--] = M[n % b];
    n /= b;
  }
  buf[MAX_BASE] = 0;
  return &buf[i] + 1;
}

int baseb_to_10(int b, const char *buf)
{
  int result = 0;
  const char *pch = NULL, *p = buf;
  int e = strlen(buf);
  while (*p)
  {
    if ((pch = strchr(M, *p)))
      result += (pch - M + 1) * (int)pow(b, --e);
    ++p;
  }
  return result;
}

int main(void)
{
  int inbase, outbase;
  char input_n[256];
  char *pch = NULL;
  char outbuf[MAX_BASE] = { 0 };

  inbase = '3';
  strcpy(input_n, "31");
  outbase = '1';

  if (!(pch = strchr(M, inbase))) exit(1);
  inbase = pch - M + 1;

  if (!(pch = strchr(M, outbase))) exit(1);
  outbase = pch - M + 1;

  puts(base10_to_b(outbase, baseb_to_10(inbase, input_n), outbuf));

  exit(0);
}

Left out user input for the values as I didn't see it as trivial to the main code itself.