Dado un entero x, escribe una función que multiplique x por 3,5 y devuelva el resultado entero. No está permitido usar %, /, *.
Examples : Input: 2 Output: 7 Input: 5 Output: 17 (Ignore the digits after decimal point)
Solución:
1. Podemos obtener x*3.5 sumando 2*x, x y x/2. Para calcular 2*x, desplace x a la izquierda en 1 y para calcular x/2, desplace x a la derecha en 1.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to multiply
// a number with 3.5
#include <iostream>
using namespace std;
int multiplyWith3Point5(int x)
{
return (x<<1) + x + (x>>1);
}
/* Driver program to test above functions*/
int main()
{
int x = 4;
cout << " "<< multiplyWith3Point5(x);
getchar();
return 0;
}
// This code is contributed by shivanisinghss2110.
C
// C++ program to multiply
// a number with 3.5
#include <stdio.h>
int multiplyWith3Point5(int x)
{
return (x<<1) + x + (x>>1);
}
/* Driver program to test above functions*/
int main()
{
int x = 4;
printf("%d", multiplyWith3Point5(x));
getchar();
return 0;
}
Java
// Java Program to multiply
// a number with 3.5
class GFG {
static int multiplyWith3Point5(int x)
{
return (x<<1) + x + (x>>1);
}
/* Driver program to test above functions*/
public static void main(String[] args)
{
int x = 2;
System.out.println(multiplyWith3Point5(x));
}
}
// This code is contributed by prerna saini.
Python3
# Python 3 program to multiply # a number with 3.5 def multiplyWith3Point5(x): return (x<<1) + x + (x>>1) # Driver program to # test above functions x = 4 print(multiplyWith3Point5(x)) # This code is contributed by # Smitha Dinesh Semwal
C#
// C# Program to multiply
// a number with 3.5
using System;
class GFG
{
static int multiplyWith3Point5(int x)
{
return (x<<1) + x + (x>>1);
}
/* Driver program to test above functions*/
public static void Main()
{
int x = 2;
Console.Write(multiplyWith3Point5(x));
}
}
// This code is contributed by Sam007
PHP
<?php
// PHP program to multiply
// a number with 3.5
function multiplyWith3Point5( $x)
{
return ($x << 1) + $x + ($x >> 1);
}
// Driver Code
$x = 4;
echo multiplyWith3Point5($x);
// This code is contributed by vt_m.
?>
Javascript
<script>
// javascript Program to multiply
// a number with 3.5
function multiplyWith3Point5(x)
{
return (x<<1) + x + (x>>1);
}
/* Driver program to test above functions*/
var x = 4;
document.write(multiplyWith3Point5(x));
// This code is contributed by Amit Katiyar
</script>
Producción
14
Complejidad de tiempo : O(1)
Complejidad espacial : O(1)
2. Otra forma de hacer esto podría ser (8*x – x)/2 (ver el código a continuación). Gracias a Ajaym por sugerir esto.
C++
// C++ program approach
#include <iostream>
using namespace std;
int multiplyWith3Point5(int x)
{
return ((x<<3) - x)>>1;
}
// This code is contributed by shivanisinghss2110
C
#include <stdio.h>
int multiplyWith3Point5(int x)
{
return ((x<<3) - x)>>1;
}
Java
// Java program for above approach
import java.io.*;
class GFG
{
// Function
static int multiplyWith3Point5(int x)
{
return ((x<<3) - x)>>1;
}
}
// This code is contributed by shivanisinghss2110
Python3
# Python program for above approach # Function def multiplyWith3Point5(x): return ((x<<3) - x)>>1 # This code is contributed by shivanisinghss2110
C#
// C# program for above approach
using System;
class GFG{
// Function to multiple number
// with 3.5
static int multiplyWith3Point5(int x)
{
return ((x<<3) - x)>>1;
}
}
// This code is contributed by shivanisinghss2110.
Javascript
// JavaScript program for above approach
// Function
function multiplyWith3Point5(x)
{
return ((x<<3) - x)>>1;
}
// This code is contributed by shivanisinghss2110
Complejidad de tiempo : O(1)
Complejidad espacial : O(1)
Otro enfoque:
Otra forma de hacer esto podría ser haciendo una multiplicación binaria por 7 y luego dividir por 2 usando solo <<, ^, & y >>.
Pero aquí tenemos que mencionar que solo se pueden pasar números positivos a este método.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for above approach
#include <iostream>
using namespace std;
// Function to multiple number
// with 3.5
int multiplyWith3Point5(int x)
{
int r = 0;
// The 3.5 is 7/2, so multiply by 7 (x * 7) then divide
// the result by 2 (result/2) x * 7 -> 7 is 0111 so by
// doing multiply by 7 it means we do 2 shifting for
// the number but since we doing multiply we need to
// take care of carry one.
int x1Shift = x << 1;
int x2Shifts = x << 2;
r = (x ^ x1Shift) ^ x2Shifts;
int c = (x & x1Shift) | (x & x2Shifts) | (x1Shift & x2Shifts);
while (c > 0) {
c <<= 1;
int t = r;
r ^= c;
c &= t;
}
// Then divide by 2
// r / 2
r = r >> 1;
return r;
}
// Driver Code
int main()
{
cout << (multiplyWith3Point5(5));
return 0;
}
// This code is contributed by Sania Kumari Gupta (kriSania804)
C
// C program for above approach
#include <stdio.h>
// Function to multiple number
// with 3.5
int multiplyWith3Point5(int x)
{
int r = 0;
// The 3.5 is 7/2, so multiply by 7 (x * 7) then divide
// the result by 2 (result/2) x * 7 -> 7 is 0111 so by
// doing multiply by 7 it means we do 2 shifting for
// the number but since we doing multiply we need to
// take care of carry one.
int x1Shift = x << 1;
int x2Shifts = x << 2;
r = (x ^ x1Shift) ^ x2Shifts;
int c = (x & x1Shift) | (x & x2Shifts) | (x1Shift & x2Shifts);
while (c > 0) {
c <<= 1;
int t = r;
r ^= c;
c &= t;
}
// Then divide by 2
// r / 2
r = r >> 1;
return r;
}
// Driver Code
int main()
{
printf("%d",(multiplyWith3Point5(5)));
return 0;
}
// This code is contributed by Sania Kumari Gupta (kriSania804)
Java
// Java program for above approach
import java.io.*;
class GFG
{
// Function to multiple number
// with 3.5
static int multiplyWith3Point5(int x)
{
int r = 0;
// The 3.5 is 7/2, so multiply
// by 7 (x * 7) then
// divide the result by 2
// (result/2) x * 7 -> 7 is
// 0111 so by doing multiply
// by 7 it means we do 2
// shifting for the number
// but since we doing
// multiply we need to take
// care of carry one.
int x1Shift = x << 1;
int x2Shifts = x << 2;
r = (x ^ x1Shift) ^ x2Shifts;
int c = (x & x1Shift) | (x & x2Shifts)
| (x1Shift & x2Shifts);
while (c > 0) {
c <<= 1;
int t = r;
r ^= c;
c &= t;
}
// Then divide by 2
// r / 2
r = r >> 1;
return r;
}
// Driver Code
public static void main(String[] args)
{
System.out.println(multiplyWith3Point5(5));
}
}
Python3
# Python3 program for the above approach # Function to multiple number # with 3.5 def multiplyWith3Point5(x): r = 0 # The 3.5 is 7/2, so multiply # by 7 (x * 7) then # divide the result by 2 # (result/2) x * 7 -> 7 is # 0111 so by doing multiply # by 7 it means we do 2 # shifting for the number # but since we doing # multiply we need to take # care of carry one. x1Shift = x << 1 x2Shifts = x << 2 r = (x ^ x1Shift) ^ x2Shifts c = (x & x1Shift) | (x & x2Shifts) | (x1Shift & x2Shifts) while (c > 0): c <<= 1 t = r r ^= c c &= t # Then divide by 2 # r / 2 r = r >> 1 return r # Driver Code if __name__ == '__main__': print(multiplyWith3Point5(5)) # This code is contributed by nirajgusain5
C#
// C# program for above approach
using System;
class GFG{
// Function to multiple number
// with 3.5
static int multiplyWith3Point5(int x)
{
int r = 0;
// The 3.5 is 7/2, so multiply
// by 7 (x * 7) then
// divide the result by 2
// (result/2) x * 7 -> 7 is
// 0111 so by doing multiply
// by 7 it means we do 2
// shifting for the number
// but since we doing
// multiply we need to take
// care of carry one.
int x1Shift = x << 1;
int x2Shifts = x << 2;
r = (x ^ x1Shift) ^ x2Shifts;
int c = (x & x1Shift) | (x & x2Shifts) |
(x1Shift & x2Shifts);
while (c > 0)
{
c <<= 1;
int t = r;
r ^= c;
c &= t;
}
// Then divide by 2
// r / 2
r = r >> 1;
return r;
}
// Driver Code
public static void Main(string[] args)
{
Console.WriteLine(multiplyWith3Point5(5));
}
}
// This code is contributed by ukasp
Javascript
<script>
// Javascript program for the above approach
// Function to multiple number
// with 3.5
function multiplyWith3Polet5(x)
{
let r = 0;
// The 3.5 is 7/2, so multiply
// by 7 (x * 7) then
// divide the result by 2
// (result/2) x * 7 -> 7 is
// 0111 so by doing multiply
// by 7 it means we do 2
// shifting for the number
// but since we doing
// multiply we need to take
// care of carry one.
let x1Shift = x << 1;
let x2Shifts = x << 2;
r = (x ^ x1Shift) ^ x2Shifts;
let c = (x & x1Shift) | (x & x2Shifts) |
(x1Shift & x2Shifts);
while (c > 0)
{
c <<= 1;
let t = r;
r ^= c;
c &= t;
}
// Then divide by 2
// r / 2
r = r >> 1;
return r;
}
// Driver code
document.write(multiplyWith3Polet5(5));
// This code is contributed by avijitmondal1998
</script>
Producción
17
Complejidad de tiempo: O (log n)
Complejidad espacial : O(1)
Escriba comentarios si encuentra que el código/algoritmo anterior es incorrecto o encuentra mejores formas de resolver el mismo problema
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA