Dados dos enteros sin signo (la máxima entrada posible puede ser de 32 bits). La tarea es sumar dos números usando operaciones con bits.
Ejemplos:
Input: n1 = 12, n2 = 34 Output: 46 Input: n1 = 12564 n2 = -1 Output: 12563
Enfoque: Como sabemos que además de poco
- 1+0=1
- 0+1=1
- 0+0=0
- 1+1=0 llevar 1
- if(llevar==1) 1+1=1 llevar 1
Representa un número entero usando la función de conjunto de bits en C++. Se comporta como una array que almacena el LSB (bit menos significativo) en el índice 0 y cuando imprimimos dicha array, imprime la representación binaria en formato inverso. Agregue cada bit desde la derecha de acuerdo con la propiedad de adición de bits y guárdelo en el tercer conjunto de bits. La función to_ulong() se usa para convertir una forma de conjunto de bits en su forma decimal.
A continuación se muestra la implementación del enfoque anterior.
C++
#include <bits/stdc++.h>
#define M 32
using namespace std;
// Function to add two bitset
int binAdd(bitset<M> atemp, bitset<M> btemp)
{
// To store the bits of answer
bitset<M> ctemp;
for (int i = 0; i < M; i++)
ctemp[i] = 0;
// Initialize carry to 0
int carry = 0;
for (int i = 0; i < M; i++) {
// Both bits are zero
if (atemp[i] + btemp[i] == 0) {
if (carry == 0)
ctemp[i] = 0;
else {
ctemp[i] = 1;
carry = 0;
}
}
// Any of the one bit is 1
else if (atemp[i] + btemp[i] == 1) {
if (carry == 0)
ctemp[i] = 1;
else {
ctemp[i] = 0;
}
}
// Both bits are 1
else {
if (carry == 0) {
ctemp[i] = 0;
carry = 1;
}
else {
ctemp[i] = 1;
}
}
}
// To convert bitset into
// decimal equivalent
return ctemp.to_ulong();
}
// Driver Code
int main()
{
int number1, number2;
number1 = 12;
number2 = 34;
// Converting number 1 to bitset form
bitset<M> num1(number1);
// Converting number 2 to bitset form
bitset<M> num2(number2);
cout << binAdd(num1, num2) << endl;
}
Java
// Java program to add two unsigned numbers using bits
import java.util.*; // Needed for BitSet class.
class GFG {
static final int M = 32;
// Function to add two BitSets. Returns long number.
static long binAdd(BitSet atemp, BitSet btemp)
{
// Initialize a 3rd BitSet to store the answer.
BitSet ctemp = new BitSet(M);
for (int i = 0; i < M; i++) {
ctemp.set(i, false);
}
// Initialize carry to ZERO.
int carry = 0;
for (int i = 0; i < M; i++) {
// Both of the bits are ZERO.
if (atemp.get(i) == false
&& // Uses the "conditional-AND" operator
// (also known as "short-circuit AND")
// which is more efficient than the
// ordinary & bitwise operator.
btemp.get(i) == false) {
if (carry == 0) {
ctemp.set(i, false);
}
else {
ctemp.set(i, true);
carry = 0;
}
}
// Either of the bits is a ONE but not both.
else if (atemp.get(i) == true
^ // Uses the "XOR" (exclusive-OR)
// operator to ensure that ONLY
// one of the two bits is a ONE.
btemp.get(i) == true) {
if (carry == 0) {
ctemp.set(i, true);
}
else {
ctemp.set(i, false);
}
}
// Both of the bits are ONE.
else // By process of elimination we do not need
// to code for the (TRUE & TRUE) condition.
{
if (carry == 0) {
ctemp.set(i, false);
carry = 1;
}
else {
ctemp.set(i, true);
}
}
}
// Returns the 3rd BitSet as its decimal equivalent
// number in long format.
return ctemp.toLongArray()[0];
}
// Driver Code
public static void main(String args[])
{
int number1, number2;
// binary literal for decimal 15.
number1 = 15;
// binary literal for decimal 85.
number2 = 1;
// result should be decimal 100.
// Converting number1 to BitSet form.
BitSet num1
= BitSet.valueOf(new long[] { number1 });
// Converting number2 to BitSet form.
BitSet num2
= BitSet.valueOf(new long[] { number2 });
System.out.printf("%d plus %d equals %d", number1,
number2, binAdd(num1, num2));
System.out.println();
}
}
// This code is contributed by Arnab Kundu.
// Modified by Matt Ambrose.
Python3
# Python3 implementation of the approach # Function to convert given Integer # to list of bits of length M def bitset(num): return [int(x) for x in format(num, '032b')] # Function to add two bitset def binAdd(atemp, btemp): # To store the bits of answer ctemp = [0] * M # Initialize carry to 0 carry = 0 for i in range(0, M): # Both bits are zero if atemp[i] + btemp[i] == 0: if carry == 0: ctemp[i] = 0 else: ctemp[i] = 1 carry = 0 # Any of the one bit is 1 elif atemp[i] + btemp[i] == 1: if carry == 0: ctemp[i] = 1 else: ctemp[i] = 0 # Both bits are 1 else: if carry == 0: ctemp[i] = 0 carry = 1 else: ctemp[i] = 1 # To convert bitset into string and then # convert string to its decimal equivalent temp = ''.join([str(x) for x in ctemp]) return int(temp, 2) # Driver Code if __name__ == "__main__": number1, number2 = 12, 34 M = 32 # Converting number 1 to bitset form num1 = bitset(number1) # Converting number 2 to bitset form num2 = bitset(number2) print(binAdd(num1, num2)) # This code is contributed by Rituraj Jain
Producción
46