UVA Solution 389 - Basically Speaking - Solution in C, C++ | Volume 3

UVA Solution 389 - Basically Speaking - Solution in C, C++ | Volume 3

UVA Online Judge Solution 389 - Basically Speaking | Volume 3
UVA Problem Link - https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=5&page=show_problem&problem=325

Problem Name: 389 - Basically Speaking
Problem Number : UVA - 389 - Basically Speaking
Online Judge : UVA Online Judge Solution
Volume: 3
Solution Language : C, C plus plus

UVA Solution 389 - Basically Speaking Code in C, CPP:

#include <stdio.h>
#include <string.h>
int main() {
    char s[50];
    int n, m;
    while(scanf("%s %d %d", s, &n, &m) == 3) {
        int i, l = strlen(s);
        long long dec = 0, j = 1;
        for(i = l-1; i >= 0; i--) {
            if(s[i] <= '9')
                dec += (s[i]-'0')*j;
            else
                dec += (s[i]-'A'+10)*j;
            j *= n;
        }
        int ans[60] = {}, idx = -1;
        while(dec > 0)
            ans[++idx] = dec%m, dec /= m;
        if(idx >= 7) {
            puts("  ERROR");
            continue;
        }
        if(idx < 0) idx = 0;
        for(i = 6; i > idx; i--)
            printf(" ");
        for(i = idx; i >= 0; i--)
            printf("%c", ans[i] < 10 ? ans[i]+'0' : ans[i]+'A'-10);
        puts("");
    }
    return 0;
}