Dados seis enteros positivos que representan Bitwise XOR y Bitwise AND de todos los pares posibles de un triplete (a, b, c) , la tarea es encontrar el triplete.
Ejemplos:
Entrada: aXORb = 30, aANDb = 0, aXORc = 10, aANDc = 20, aXORb = 20, aANDb = 10
Salida: a = 10, b = 20, c= 30
Explicación:
Si a = 10, b = 20, c = 30
a ^ b = 30, a & b = 0
a ^ c = 10, a & c = 20
a ^ b = 20, a & b = 10
Por lo tanto, la salida requerida es (a, b, c) = ( 10, 20, 30).Entrada: aXORb = 3, aANDb = 0, aXORc = 2, aANDc = 1, aXORb = 1, aANDb = 2
Salida: a = 1, b = 2, c = 3
Enfoque: la idea es encontrar la suma de todos los pares posibles de triplete utilizando sus valores Bitwise XOR y Bitwise AND en función de las siguientes observaciones:
a + b = a ^ b + 2 * (a y b)
Siga los pasos a continuación para resolver el problema:
- Encuentre la suma de cada posible par de tripletes, es decir, (a + b, b + c, c + a) usando la fórmula anterior.
- El valor de a se puede calcular como ((a + b) + (a + c) – (b + c)) / 2 .
- El valor de b se puede calcular como ((a + b) – a) .
- El valor de c se puede calcular como ((b + c) – b) .
- Finalmente, imprima el valor del triplete (a, b, c) .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the triplet with given
// Bitwise XOR and Bitwise AND values of all
// possible pairs of the triplet
void findNumbers(int aXORb, int aANDb, int aXORc, int aANDc,
int bXORc, int bANDc)
{
// Stores values of
// a triplet
int a, b, c;
// Stores a + b
int aSUMb;
// Stores a + c
int aSUMc;
// Stores b + c
int bSUMc;
// Calculate aSUMb
aSUMb = aXORb + aANDb * 2;
// Calculate aSUMc
aSUMc = aXORc + aANDc * 2;
// Calculate bSUMc
bSUMc = bXORc + bANDc * 2;
// Calculate a
a = (aSUMb - bSUMc + aSUMc) / 2;
// Calculate b
b = aSUMb - a;
// Calculate c
c = aSUMc - a;
// Print a
cout << "a = " << a;
// Print b
cout << ", b = " << b;
// Print c
cout << ", c = " << c;
}
// Driver Code
int main()
{
int aXORb = 30, aANDb = 0, aXORc = 20, aANDc = 10,
bXORc = 10, bANDc = 20;
findNumbers(aXORb, aANDb, aXORc, aANDc, bXORc, bANDc);
}
Java
// Java program to implement
// the above approach
class GFG{
// Function to find the triplet with given
// Bitwise XOR and Bitwise AND values of all
// possible pairs of the triplet
static void findNumbers(int aXORb, int aANDb,
int aXORc, int aANDc,
int bXORc, int bANDc)
{
// Stores values of
// a triplet
int a, b, c;
// Stores a + b
int aSUMb;
// Stores a + c
int aSUMc;
// Stores b + c
int bSUMc;
// Calculate aSUMb
aSUMb = aXORb + aANDb * 2;
// Calculate aSUMc
aSUMc = aXORc + aANDc * 2;
// Calculate bSUMc
bSUMc = bXORc + bANDc * 2;
// Calculate a
a = (aSUMb - bSUMc + aSUMc) / 2;
// Calculate b
b = aSUMb - a;
// Calculate c
c = aSUMc - a;
// Print a
System.out.print("a = " + a);
// Print b
System.out.print(", b = " + b);
// Print c
System.out.print(", c = " + c);
}
// Driver Code
public static void main(String[] args)
{
int aXORb = 30, aANDb = 0,
aXORc = 20, aANDc = 10,
bXORc = 10, bANDc = 20;
findNumbers(aXORb, aANDb, aXORc,
aANDc, bXORc, bANDc);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python program to implement
# the above approach
# Function to find the triplet with given
# Bitwise XOR and Bitwise AND values of all
# possible pairs of the triplet
def findNumbers(aXORb, aANDb, aXORc, aANDc, bXORc, bANDc):
# Stores values of
# a triplet
a, b, c = 0, 0, 0;
# Stores a + b
aSUMb = 0;
# Stores a + c
aSUMc = 0;
# Stores b + c
bSUMc = 0;
# Calculate aSUMb
aSUMb = aXORb + aANDb * 2;
# Calculate aSUMc
aSUMc = aXORc + aANDc * 2;
# Calculate bSUMc
bSUMc = bXORc + bANDc * 2;
# Calculate a
a = (aSUMb - bSUMc + aSUMc) // 2;
# Calculate b
b = aSUMb - a;
# Calculate c
c = aSUMc - a;
# Pra
print("a = " , a, end = "");
# Prb
print(", b = " , b, end = "");
# Prc
print(", c = " , c, end = "");
# Driver Code
if __name__ == '__main__':
aXORb = 30; aANDb = 0; aXORc = 20; aANDc = 10; bXORc = 10; bANDc = 20;
findNumbers(aXORb, aANDb, aXORc, aANDc, bXORc, bANDc);
# This code contributed by shikhasingrajput
C#
// C# code for above approach
using System;
public class GFG
{
// Function to find the triplet with given
// Bitwise XOR and Bitwise AND values of all
// possible pairs of the triplet
static void findNumbers(int aXORb, int aANDb,
int aXORc, int aANDc,
int bXORc, int bANDc)
{
// Stores values of
// a triplet
int a, b, c;
// Stores a + b
int aSUMb;
// Stores a + c
int aSUMc;
// Stores b + c
int bSUMc;
// Calculate aSUMb
aSUMb = aXORb + aANDb * 2;
// Calculate aSUMc
aSUMc = aXORc + aANDc * 2;
// Calculate bSUMc
bSUMc = bXORc + bANDc * 2;
// Calculate a
a = (aSUMb - bSUMc + aSUMc) / 2;
// Calculate b
b = aSUMb - a;
// Calculate c
c = aSUMc - a;
// Print a
System.Console.Write("a = " + a);
// Print b
System.Console.Write(", b = " + b);
// Print c
System.Console.Write(", c = " + c);
}
// Driver code
static public void Main ()
{
int aXORb = 30, aANDb = 0,
aXORc = 20, aANDc = 10,
bXORc = 10, bANDc = 20;
findNumbers(aXORb, aANDb, aXORc,
aANDc, bXORc, bANDc);
}
}
// This code is contributed by offbeat.
Javascript
<script>
// JavaScript program to implement
// the above approach
// Function to find the triplet with given
// Bitwise XOR and Bitwise AND values of all
// possible pairs of the triplet
function findNumbers(aXORb, aANDb, aXORc, aANDc, bXORc, bANDc)
{
// Stores values of
// a triplet
let a, b, c;
// Stores a + b
let aSUMb;
// Stores a + c
let aSUMc;
// Stores b + c
let bSUMc;
// Calculate aSUMb
aSUMb = aXORb + aANDb * 2;
// Calculate aSUMc
aSUMc = aXORc + aANDc * 2;
// Calculate bSUMc
bSUMc = bXORc + bANDc * 2;
// Calculate a
a = Math.floor((aSUMb - bSUMc + aSUMc) / 2);
// Calculate b
b = aSUMb - a;
// Calculate c
c = aSUMc - a;
// Print a
document.write("a = " + a);
// Print b
document.write(", b = " + b);
// Print c
document.write(", c = " + c);
}
// Driver Code
let aXORb = 30, aANDb = 0, aXORc = 20, aANDc = 10,
bXORc = 10, bANDc = 20;
findNumbers(aXORb, aANDb, aXORc, aANDc, bXORc, bANDc);
// This code is contributed by Surbhi Tyagi.
</script>
Producción:
a = 10, b = 20, c = 30
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)