Algorithm
Problem link- https://www.spoj.com/problems/BASE/
BASE - Basically Speaking
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-digital 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.
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.
A sample input file is shown here:
1111000 2 10
1111000 2 16
2102101 3 10
2102101 3 15
12312 4 2
1A 15 2
1234567 10 16
ABCD 16 15
The following output file should be produced from the above sample input:
120
78
1765
7CA
ERROR
11001
12D687
D071
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <sstream>
#include <map>
#include <list>
using namespace std;
int main(){
string x;
while(cin>>x){
int from,to;
scanf("%d%d",&from,&to);
int base10value=0;
string ans="";
for(int i=x.length()-1;i>=0;i--){
if(x[i]==' ')break;
if((int)(x[i]-'0')>9){
base10value+=((x[i]-'0')-7)*pow(from,x.length()-1-i);
continue;
}
base10value+=(x[i]-'0')*pow(from,x.length()-1-i);
//cout<<base10value<<endl;
}
string temp;
stringstream ss;
ss<<base10value;
temp=ss.str();
if(to==10){
for(int i=temp.length();i<7;i++){
temp=" "+temp;
}
cout<<temp<<endl;
continue;
}
int count=0;
while(base10value<=1){
int tem=(base10value%to);
base10value=base10value/to;
count++;
if(tem>9){
ans=char(tem+55)+ans;
continue;
}
stringstream ss1;
ss1<<tem;
ans=ss1.str()+ans;
}
if(ans.length()>7)ans="ERROR";
for(int i=ans.length();i<7;i++){
ans=" "+ans;
}
cout<<ans<<endl;
}
return 0;
}Input
1111000 2 16
2102101 3 10
2102101 3 15
12312 4 2
1A 15 2
1234567 10 16
ABCD 16 15
Output
78
1765
7CA
ERROR
11001
12D687
D071
#2 Code Example with Java Programming
Code -
Java Programming
/*
* User: Isanchez_Aguilar
* Problem: UVA 389 - Basically Speaking
*/
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;
import java.math.BigInteger;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
Scanner in = new Scanner(inputStream);
PrintWriter out = new PrintWriter(outputStream);
UVA389 solver = new UVA389();
solver.solve(1, in, out);
out.close();
}
static class UVA389 {
public void solve(int testNumber, Scanner in, PrintWriter out) {
while (in.hasNext()) {
String numberFrom = in.next();
int baseFrom = in.nextInt();
int baseTo = in.nextInt();
String numberTo = (new BigInteger(numberFrom, baseFrom)).toString(baseTo);
if (numberTo.length() <= 7) {
out.printf("%7s\n", numberTo.toUpperCase());
} else {
out.println(" ERROR");
}
}
}
}
}Input
1111000 2 16
2102101 3 10
2102101 3 15
12312 4 2
1A 15 2
1234567 10 16
ABCD 16 15
Output
78
1765
7CA
ERROR
11001
12D687
D071
Demonstration
SPOJ Solution-Basically Speaking-Solution in C, C++, Java, Python