Basically Speaking

Basically Speaking

Time Limit: 2 Sec  Memory Limit: 200 MB

Submit: 19  Solved: 11

[Submit][Status][Web Board]

Description

The Really Neato Calculator Company, Inc. has recently hired your team to help design

their Super Neato Model I calculator. As a computer scientist you suggested to the

company that it would be neato if this new calculator could convert among number bases.

The company thought this was a stupendous idea and has asked your team to come up

with the prototype program for doing base conversion. The project manager of the Super

Neato Model I calculator has informed you that the calculator will have the following neato features:

It will have a 7-digit display.

Its buttons will include the capital letters A through F in addition to the digits 0 through 9.

It will support bases 2 through 16.

Input

The input for your prototype program will consist of one base conversion per line.

There will be three numbers per line. The first number will be the number in the

base you are converting from. The second number is the base you are converting from.

The third number is the base you are converting to. There will be one or more blanks

surrounding (on either side of) the numbers. There are several lines of input and your

program should continue to read until the end of file is reached.

Output

The output will only be the converted number as it would appear on the display of the

calculator. The number should be right justified in the 7-digit display. If the number is to large to appear on the display,

then print ``ERROR'' (without the quotes) right justified in the display.

Sample Input

1111000 2 10

1111000 2 16

2102101  3 10

2102101 3   15

12312 4   2

1A   15 2

1234567 10 16

ABCD 16 15

Sample Output

120

78

1765

7CA

ERROR

11001

12D687

D071

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a,b,l,i;
long long sum;
char ch[];
int work(char ch)
{
    if(ch>='' && ch<='') return ch-'';
     return ch-'A'+;
}
char solve(int k)
{
    if (k<) return char(k+);
      return char(k+);
}
int main()
{
    while(~scanf("%s%d%d",&ch,&a,&b))
    {
        l=strlen(ch);
        sum=;
        for(int i=;i<l;i++)
            sum=sum*a+work(ch[i]);
        if (sum==)
        {
            printf("      0\n");
            continue;
        }
        l=;
        while(sum>)
        {
            ch[++l]=solve(sum%b);
            sum=sum/b;
        }
       if (l<=)
        {
            for(i=;i<=-l;i++) printf(" ");
            for(i=l;i>=;i--) printf("%c",ch[i]);
        }
        else printf("  ERROR");
       printf("\n");
    }
    return ;
}