给定数字n,编写一个函数,如果n被9整除,则该函数返回true,否则返回false。检查n除以9的最简单方法是执行n%9。
另一种方法是对n的数字求和。如果数字总和是9的倍数,则n是9的倍数。
以上方法不是基于位运算运算符的方法,需要使用%和/。
位运算运算符通常比模和除法运算符快。以下是一种基于位运算运算符的方法,用于检查除数是否为9。
C++
// C++ program to check if a number
// is multiple of 9 using bitwise operators
#include
using namespace std;
// Bitwise operator based function to check divisibility by 9
bool isDivBy9(int n)
{
// Base cases
if (n == 0 || n == 9)
return true;
if (n < 9)
return false;
// If n is greater than 9, then recur for [floor(n/9) - n%8]
return isDivBy9((int)(n >> 3) - (int)(n & 7));
}
// Driver program to test above function
int main()
{
// Let us print all multiples of 9 from 0 to 100
// using above method
for (int i = 0; i < 100; i++)
if (isDivBy9(i))
cout << i << " ";
return 0;
}
Java
// Java program to check if a number
// is multiple of 9 using bitwise operators
import java.lang.*;
class GFG {
// Bitwise operator based function
// to check divisibility by 9
static boolean isDivBy9(int n)
{
// Base cases
if (n == 0 || n == 9)
return true;
if (n < 9)
return false;
// If n is greater than 9, then
// recur for [floor(n/9) - n%8]
return isDivBy9((int)(n >> 3) - (int)(n & 7));
}
// Driver code
public static void main(String arg[])
{
// Let us print all multiples of 9 from
// 0 to 100 using above method
for (int i = 0; i < 100; i++)
if (isDivBy9(i))
System.out.print(i + " ");
}
}
// This code is contributed by Anant Agarwal.
Python3
# Bitwise operator based
# function to check divisibility by 9
def isDivBy9(n):
# Base cases
if (n == 0 or n == 9):
return True
if (n < 9):
return False
# If n is greater than 9,
# then recur for [floor(n / 9) - n % 8]
return isDivBy9((int)(n>>3) - (int)(n&7))
# Driver code
# Let us print all multiples
# of 9 from 0 to 100
# using above method
for i in range(100):
if (isDivBy9(i)):
print(i, " ", end ="")
# This code is contributed
# by Anant Agarwal.
C#
// C# program to check if a number
// is multiple of 9 using bitwise operators
using System;
class GFG {
// Bitwise operator based function
// to check divisibility by 9
static bool isDivBy9(int n)
{
// Base cases
if (n == 0 || n == 9)
return true;
if (n < 9)
return false;
// If n is greater than 9, then
// recur for [floor(n/9) - n%8]
return isDivBy9((int)(n >> 3) - (int)(n & 7));
}
// Driver code
public static void Main()
{
// Let us print all multiples of 9 from
// 0 to 100 using above method
for (int i = 0; i < 100; i++)
if (isDivBy9(i))
Console.Write(i + " ");
}
}
// This code is contributed by nitin mittal.
PHP
> 3) -
($n & 7));
}
// Driver Code
// Let us print all multiples
// of 9 from 0 to 100
// using above method
for ($i = 0; $i < 100; $i++)
if (isDivBy9($i))
echo $i ," ";
// This code is contributed by nitin mittal
?>
Javascript
输出:
0 9 18 27 36 45 54 63 72 81 90 99
这是如何运作的?
n / 9可以使用以下简单公式以n / 8的形式编写。
n/9 = n/8 - n/72
由于我们需要使用按位运算运算符,因此我们使用n >> 3获得floor(n / 8)的值,并使用n&7获得n%8的值。我们需要用floor(n / 8)和n%8来表示上面的表达式。
n / 8等于“ floor(n / 8)+(n%8)/ 8” 。让我们用floor(n / 8)和n%8来写上面的表达式
n/9 = floor(n/8) + (n%8)/8 - [floor(n/8) + (n%8)/8]/9
n/9 = floor(n/8) - [floor(n/8) - 9(n%8)/8 + (n%8)/8]/9
n/9 = floor(n/8) - [floor(n/8) - n%8]/9