给定整数x,编写一个将x乘以3.5并返回整数结果的函数。不允许使用%,/,*。
Examples :
Input: 2
Output: 7
Input: 5
Output: 17 (Ignore the digits after decimal point)
解决方案:
1.我们可以通过将2 * x,x和x / 2相加得到x * 3.5。要计算2 * x,请将x左移1,并计算x / 2,x右移2。
下面是上述方法的实现:
C++
// C++ program to multiply
// a number with 3.5
#include
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
> 1);
}
// Driver Code
$x = 4;
echo multiplyWith3Point5($x);
// This code is contributed by vt_m.
?>
Javascript
C
#include
int multiplyWith3Point5(int x)
{
return ((x<<3) - x)>>1;
}
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 mutiply
// 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));
}
}
输出
14
2.执行此操作的另一种方法是(8 * x – x)/ 2(请参见下面的代码)。感谢Ajaym提出的建议。
C
#include
int multiplyWith3Point5(int x)
{
return ((x<<3) - x)>>1;
}
另一种方法:
另一种方法是将二进制乘以7,然后仅使用<<,^,&和>>将其除以2。
但是在这里我们不得不提到,只有正数可以传递给该方法。
下面是上述方法的实现:
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 mutiply
// 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));
}
}
输出
17