我们想到的第一个解决方案是我们在学校中学到的解决方案。如果数字中的数字总和是3的倍数，则数字是3的倍数，例如，对于612，数字的总和是9，因此它是3的倍数。但是这种解决方案效率不高。您必须一一获取所有十进制数字，将它们相加，然后检查sum是否为3的倍数。
数字的二进制表示形式中有一个模式，可用于确定数字是否为3的倍数。如果奇数设置位(在奇数位置设置的位)与偶数设置位的计数之间的差为3的倍数，则为数字。
范例： 23(00..10111)
1)获取奇数位置所有置位的计数(对于23，为3)。
2)获取偶数位置的所有设置位的计数(对于23，则为1)。
3)如果上述两个计数的差是3的倍数，则数字也是3的倍数。
(对于23，它是2，所以23不是3的倍数)
再举一些例子，例如21、15等…
Algorithm: isMutlipleOf3(n)
1) Make n positive if n is negative.
2) If number is 0 then return 1
3) If number is 1 then return 0
4) Initialize: odd_count = 0, even_count = 0
5) Loop while n != 0
a) If rightmost bit is set then increment odd count.
b) Right-shift n by 1 bit
c) If rightmost bit is set then increment even count.
d) Right-shift n by 1 bit
6) return isMutlipleOf3(odd_count - even_count)证明：
通过以十进制数字11为例可以证明以上内容。 (在这种情况下，十进制数字中的11与二进制数字中的3相同)
如果奇数和偶数之和之间的差是11的倍数，那么十进制数是11的倍数。让我们看看如何。
让我们以十进制的2位数字为例
AB = 11A -A + B = 11A +(B – A)
因此，如果(B – A)是11的倍数，则AB是。
让我们以3位数字表示。
ABC = 99A + A + 11B – B + C =(99A + 11B)+(A + C – B)
因此，如果(A + C – B)为11的倍数，则为(ABC)
现在让我们取4位数字。
ABCD = 1001A + D + 11C – C + 999B + B – A
=(1001A – 999B + 11C)+(D + B – A -C)
因此，如果(B + D – A – C)是11的倍数，则ABCD为ABCD。
可以对所有十进制数字继续执行此操作。
上面的概念可以用相同的方式证明为3的二进制数。
时间复杂度： O(logn)
程序：
C++
// CPP program to check if n is a multiple of 3
#include
using namespace std;
/* Function to check if n is a multiple of 3*/
int isMultipleOf3(int n)
{
int odd_count = 0;
int even_count = 0;
/* Make no positive if +n is multiple of 3
then is -n. We are doing this to avoid
stack overflow in recursion*/
if (n < 0)
n = -n;
if (n == 0)
return 1;
if (n == 1)
return 0;
while (n) {
/* If odd bit is set then
increment odd counter */
if (n & 1)
odd_count++;
/* If even bit is set then
increment even counter */
if (n & 2)
even_count++;
n = n >> 2;
}
return isMultipleOf3(abs(odd_count - even_count));
}
/* Program to test function isMultipleOf3 */
int main()
{
int num = 24;
if (isMultipleOf3(num))
printf("%d is multiple of 3", num);
else
printf("%d is not a multiple of 3", num);
return 0;
} Java
// Java program to check if
// n is a multiple of 3
import java.lang.*;
import java.util.*;
class GFG {
// Function to check if n
// is a multiple of 3
static int isMultipleOf3(int n)
{
int odd_count = 0;
int even_count = 0;
/* Make no positive if +n is multiple
of 3 then is -n. We are doing this to
avoid stack overflow in recursion*/
if (n < 0)
n = -n;
if (n == 0)
return 1;
if (n == 1)
return 0;
while (n != 0) {
/* If odd bit is set then
increment odd counter */
if ((n & 1) != 0)
odd_count++;
/* If even bit is set then
increment even counter */
if ((n & 2) != 0)
even_count++;
n = n >> 2;
}
return isMultipleOf3(Math.abs(odd_count - even_count));
}
/* Program to test function isMultipleOf3 */
public static void main(String[] args)
{
int num = 24;
if (isMultipleOf3(num) != 0)
System.out.println(num + " is multiple of 3");
else
System.out.println(num + " is not a multiple of 3");
}
}
// This code is contributed by Sahil_BansallPython3
# Python profram to check if n is a multiple of 3
# Function to check if n is a multiple of 3
def isMultipleOf3(n):
odd_count = 0
even_count = 0
# Make no positive if + n is multiple of 3
# then is -n. We are doing this to avoid
# stack overflow in recursion
if(n < 0):
n = -n
if(n == 0):
return 1
if(n == 1):
return 0
while(n):
# If odd bit is set then
# increment odd counter
if(n & 1):
odd_count += 1
# If even bit is set then
# increment even counter
if(n & 2):
even_count += 1
n = n >> 2
return isMultipleOf3(abs(odd_count - even_count))
# Program to test function isMultipleOf3
num = 24
if (isMultipleOf3(num)):
print(num, 'is multiple of 3')
else:
print(num, 'is not a multiple of 3')
# This code is contributed by Danish RazaC#
// C# program to check if
// n is a multiple of 3
using System;
class GFG {
// Function to check if n
// is a multiple of 3
static int isMultipleOf3(int n)
{
int odd_count = 0, even_count = 0;
/* Make no positive if +n is multiple
of 3 then is -n. We are doing this to
avoid stack overflow in recursion*/
if (n < 0)
n = -n;
if (n == 0)
return 1;
if (n == 1)
return 0;
while (n != 0) {
/* If odd bit is set then
increment odd counter */
if ((n & 1) != 0)
odd_count++;
/* If even bit is set then
increment even counter */
if ((n & 2) != 0)
even_count++;
n = n >> 2;
}
return isMultipleOf3(Math.Abs(odd_count - even_count));
}
// Driver Code
public static void Main()
{
int num = 24;
if (isMultipleOf3(num) != 0)
Console.Write(num + " is multiple of 3");
else
Console.Write(num + " is not a multiple of 3");
}
}
// This code is contributed by Sam007PHP
> 2;
}
return isMultipleOf3(abs($odd_count -
$even_count));
}
// Driver Code
$num = 24;
if (isMultipleOf3($num))
echo $num, "is multiple of 3";
else
echo $num, "is not a multiple of 3";
// This code is contributed by vt_m.
?>Javascript
输出 ：
24 is multiple of 3