不使用算术运算符检查 n 是否能被 2 的幂整除
给定两个正整数 n 和 m 。问题是在不使用算术运算符的情况下，检查 n 是否可被 2 m 整除。
示例:
Input : n = 8, m = 2
Output : Yes
Input : n = 14, m = 3
Output : No
方法:如果一个数可以被 2 整除，那么它的最低有效位(LSB)设置为 0，如果可以被 4 整除，那么两个 LSB 设置为 0，如果可以被 8 整除，那么三个 LSB 设置为 0，以此类推。记住这一点，如果(n&((1<<m)–1)等于 0，则数字 n 可被2mT13】整除。
c++
// C++ implementation to check whether n
// is divisible by pow(2, m)
#include <bits/stdc++.h>
using namespace std;
// function to check whether n
// is divisible by pow(2, m)
bool isDivBy2PowerM(unsigned int n,
unsigned int m)
{
// if expression results to 0, then
// n is divisible by pow(2, m)
if ((n & ((1 << m) - 1)) == 0)
return true;
// n is not divisible
return false;
}
// Driver program to test above
int main()
{
unsigned int n = 8, m = 2;
if (isDivBy2PowerM(n, m))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// JAVA Code for Check if n is divisible
// by power of 2 without using arithmetic
// operators
import java.util.*;
class GFG {
// function to check whether n
// is divisible by pow(2, m)
static boolean isDivBy2PowerM(int n,
int m)
{
// if expression results to 0, then
// n is divisible by pow(2, m)
if ((n & ((1 << m) - 1)) == 0)
return true;
// n is not divisible
return false;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 8, m = 2;
if (isDivBy2PowerM(n, m))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Arnav Kr. Mandal.
python 3
T2
c#
**T31
// C# Code for Check if n is divisible
// by power of 2 without using arithmetic
// operators
using System;
class GFG {
// function to check whether n
// is divisible by pow(2, m)
static bool isDivBy2PowerM(int n, int m)
{
// if expression results to 0, then
// n is divisible by pow(2, m)
if ((n & ((1 << m) - 1)) == 0)
return true;
// n is not divisible
return false;
}
/* Driver program to test above function */
public static void Main()
{
int n = 8, m = 2;
if (isDivBy2PowerM(n, m))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Sam007
```** **PHPT4T37】JavascriptT40**
**输出:**
Yes ```
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