在不使用除法(/)和模数(%)运算符的情况下计算n模d,其中d是2的幂。
从右起第i位设置为d。为了获得n模数d,我们只需要按原样返回0到i -1(从右)的n位,其他位返回0。
例如,如果n = 6(00..110)和d = 4(00..100)。 d中的最后一个设置位在位置3(从右侧)。因此,我们需要按原样返回n的最后两位,其他位返回0,即00..010。
现在做起来非常容易,猜出来……。
是的,您猜对了。请参阅以下程序。
C++
#include
// This function will return n % d.
// d must be one of: 1, 2, 4, 8, 16, 32, …
unsigned int getModulo(unsigned int n,
unsigned int d)
{
return ( n & (d - 1) );
}
// Driver Code
int main()
{
unsigned int n = 6;
// d must be a power of 2
unsigned int d = 4;
printf("%u moduo %u is %u", n, d, getModulo(n, d));
getchar();
return 0;
}
Java
// Java code for Compute modulus division by
// a power-of-2-number
class GFG {
// This function will return n % d.
// d must be one of: 1, 2, 4, 8, 16, 32,
static int getModulo(int n, int d)
{
return ( n & (d-1) );
}
// Driver Code
public static void main(String[] args)
{
int n = 6;
/*d must be a power of 2*/
int d = 4;
System.out.println(n+" moduo " + d +
" is " + getModulo(n, d));
}
}
// This code is contributed
// by Smitha Dinesh Semwal.
Python3
# Python code to demonstrate
# modulus division by power of 2
# This function will
# return n % d.
# d must be one of:
# 1, 2, 4, 8, 16, 32, …
def getModulo(n, d):
return ( n & (d-1) )
# Driver program to
# test above function
n = 6
#d must be a power of 2
d = 4
print(n,"moduo",d,"is",
getModulo(n, d))
# This code is contributed by
# Smitha Dinesh Semwal
C#
// C# code for Compute modulus
// division by a power-of-2-number
using System;
class GFG {
// This function will return n % d.
// d must be one of: 1, 2, 4, 8, 16, 32, …
static uint getModulo( uint n, uint d)
{
return ( n & (d-1) );
}
// Driver code
static public void Main ()
{
uint n = 6;
uint d = 4; /*d must be a power of 2*/
Console.WriteLine( n + " moduo " + d +
" is " + getModulo(n, d));
}
}
// This code is contributed by vt_m.
PHP
Javascript
https://www.youtube.com/watch?v=fSjW
-wDghTs