Dados tres enteros positivos X , Y y B , donde X e Y son enteros de base B , la tarea es encontrar el valor de X – Y tal que X >= Y .
Ejemplos:
Entrada: X = 1212, Y = 256, B = 8
Salida: 0734
Explicación: El valor de 1212 – 256 en base 8 es 734 .Entrada: X = 546, Y = 248, B = 9
Salida: 287
Enfoque: El problema dado se puede resolver mediante el uso de restas matemáticas básicas. Siga los pasos a continuación para resolver el problema dado:
- Inicialice dos variables, digamos potencia = 1 , acarreo = 0 , para realizar un seguimiento de la potencia actual y el acarreo generado mientras se resta respectivamente.
- Inicialice una variable, digamos finalVal = 0 , para almacenar el valor resultante de X – Y.
- Iterar un ciclo hasta que X > 0 y realizar los siguientes pasos:
- Almacene los últimos dígitos del valor actual de X e Y en dos variables, digamos n1 = X % 10 y n2 = Y % 10 respectivamente.
- Elimine los últimos dígitos de X e Y actualizando X = X / 10 e Y = Y / 10 .
- Inicializar temp = n1 – n2 + carry .
- Si temp < 0 , agregue base B a N , es decir N = N + B y establezca carry = -1 , que actuará como un préstamo. De lo contrario, establezca llevar = 0 .
- Agregue temp * power actual a finalVal, es decir, finalVal = finalVal + temp * power y configure power = power * 10 .
- Después de completar los pasos anteriores, imprima el valor de finalVal como resultado.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find X - Y in base B
int getDifference(int B, int X, int Y)
{
// To store final answer
int finalVal = 0;
// To store carry generated
int carry = 0;
// To keep track of power
int power = 1;
while (X > 0) {
// Store last digits of current
// value of X and Y in n1 and
// n2 respectively
int n1 = X % 10;
int n2 = Y % 10;
// Remove last digits from
// X and Y
X = X / 10;
Y = Y / 10;
int temp = n1 - n2 + carry;
if (temp < 0) {
// Carry = -1 will act
// as borrow
carry = -1;
temp += B;
}
else {
carry = 0;
}
// Add in final result
finalVal += temp * power;
power = power * 10;
}
// Return final result
return finalVal;
}
// Driver Code
int main()
{
int X = 1212;
int Y = 256;
int B = 8;
cout << (getDifference(B, X, Y));
return 0;
}
// This code is contributed by rakeshsahni
Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to find X - Y in base B
public static int getDifference(
int B, int X, int Y)
{
// To store final answer
int finalVal = 0;
// To store carry generated
int carry = 0;
// To keep track of power
int power = 1;
while (X > 0) {
// Store last digits of current
// value of X and Y in n1 and
// n2 respectively
int n1 = X % 10;
int n2 = Y % 10;
// Remove last digits from
// X and Y
X = X / 10;
Y = Y / 10;
int temp = n1 - n2 + carry;
if (temp < 0) {
// Carry = -1 will act
// as borrow
carry = -1;
temp += B;
}
else {
carry = 0;
}
// Add in final result
finalVal += temp * power;
power = power * 10;
}
// Return final result
return finalVal;
}
// Driver Code
public static void main(String[] args)
{
int X = 1212;
int Y = 256;
int B = 8;
System.out.println(
getDifference(B, X, Y));
}
}
Python3
# Python3 program for the above approach # Function to find X - Y in base B def getDifference(B, X, Y) : # To store final answer finalVal = 0; # To store carry generated carry = 0; # To keep track of power power = 1; while (X > 0) : # Store last digits of current # value of X and Y in n1 and # n2 respectively n1 = X % 10; n2 = Y % 10; # Remove last digits from # X and Y X = X // 10; Y = Y // 10; temp = n1 - n2 + carry; if (temp < 0) : # Carry = -1 will act # as borrow carry = -1; temp += B; else : carry = 0; # Add in final result finalVal += temp * power; power = power * 10; # Return final result return finalVal; # Driver Code if __name__ == "__main__" : X = 1212; Y = 256; B = 8; print(getDifference(B, X, Y)); # This code is contributed by AnkThon
C#
// C# program for the above approach
using System;
public class GFG
{
// Function to find X - Y in base B
public static int getDifference(int B, int X, int Y)
{
// To store final answer
int finalVal = 0;
// To store carry generated
int carry = 0;
// To keep track of power
int power = 1;
while (X > 0) {
// Store last digits of current
// value of X and Y in n1 and
// n2 respectively
int n1 = X % 10;
int n2 = Y % 10;
// Remove last digits from
// X and Y
X = X / 10;
Y = Y / 10;
int temp = n1 - n2 + carry;
if (temp < 0) {
// Carry = -1 will act
// as borrow
carry = -1;
temp += B;
}
else {
carry = 0;
}
// Add in final result
finalVal += temp * power;
power = power * 10;
}
// Return final result
return finalVal;
}
// Driver Code
public static void Main(string[] args)
{
int X = 1212;
int Y = 256;
int B = 8;
Console.WriteLine(getDifference(B, X, Y));
}
}
// This code is contributed by AnkThon
Javascript
<script>
// Javascript program for the above approach
// Function to find X - Y in base B
function getDifference(B, X, Y) {
// To store final answer
let finalVal = 0;
// To store carry generated
let carry = 0;
// To keep track of power
let power = 1;
while (X > 0) {
// Store last digits of current
// value of X and Y in n1 and
// n2 respectively
let n1 = X % 10;
let n2 = Y % 10;
// Remove last digits from
// X and Y
X = Math.floor(X / 10);
Y = Math.floor(Y / 10);
let temp = n1 - n2 + carry;
if (temp < 0) {
// Carry = -1 will act
// as borrow
carry = -1;
temp += B;
} else {
carry = 0;
}
// Add in final result
finalVal += temp * power;
power = power * 10;
}
// Return final result
return finalVal;
}
// Driver Code
let X = 1212;
let Y = 256;
let B = 8;
document.write(getDifference(B, X, Y));
// This code is contributed by gfgking
</script>
Producción:
734
Complejidad de tiempo: O(log 10 N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por atharvakarpe y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA