Answer:
#include <math.h>
#include <stdio.h>
#include <string.h>
#define BASE 3
#define NRQUESTIONS 15
void toABC(int n, char* buf, int base, int size) {
memset(buf, 'A', size);
buf[size] = 0;
while (n && size) {
buf[--size] = 'A' + (n % base);
n /= base;
}
}
int main()
{
char buf[16];
for (int i = 0; i < pow(BASE, NRQUESTIONS); i++) {
toABC(i, buf, BASE, NRQUESTIONS);
printf("%s\n", buf);
}
}
Explanation:
Assuming 3 is the number of possible answers to choose from for each question.
I tackled this by having an integer counter enumerate all values from 0 to 3^15, and then convert each integer to a base-3 representation, using ABC in stead of 012.
