给定两个整数(小于2 ^ 31),A和B。任务是查找二进制表示形式不同的位数。
例子:
Input : A = 12, B = 15
Output : Number of different bits : 2
Explanation: The binary representation of
12 is 1100 and 15 is 1111.
So, the number of different bits are 2.
Input : A = 3, B = 16
Output : Number of different bits : 3
方法:
- 从“ 0”到“ 31”运行一个循环,然后将A和B的位右移“ i”位,然后检查“ 0th”位置的位是否不同。
- 如果位不同,则增加计数。
- 由于数字小于2 ^ 31,因此我们只需要运行循环“ 32”次,即从“ 0”到“ 31”。
- 如果我们将数字按1与1相加,就可以得到第一位。
- 在循环末尾显示计数。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// compute number of different bits
void solve(int A, int B)
{
int count = 0;
// since, the numbers are less than 2^31
// run the loop from '0' to '31' only
for (int i = 0; i < 32; i++) {
// right shift both the numbers by 'i' and
// check if the bit at the 0th position is different
if (((A >> i) & 1) != ((B >> i) & 1)) {
count++;
}
}
cout << "Number of different bits : " << count << endl;
}
// Driver code
int main()
{
int A = 12, B = 15;
// find number of different bits
solve(A, B);
return 0;
}
Java
// Java implementation of the approach
import java.io.*;
class GFG {
// compute number of different bits
static void solve(int A, int B)
{
int count = 0;
// since, the numbers are less than 2^31
// run the loop from '0' to '31' only
for (int i = 0; i < 32; i++) {
// right shift both the numbers by 'i' and
// check if the bit at the 0th position is different
if (((A >> i) & 1) != ((B >> i) & 1)) {
count++;
}
}
System.out.println("Number of different bits : " + count);
}
// Driver code
public static void main (String[] args) {
int A = 12, B = 15;
// find number of different bits
solve(A, B);
}
}
// this code is contributed by anuj_67..
Python3
# Python3 implementation of the approach
# compute number of different bits
def solve( A, B):
count = 0
# since, the numbers are less than 2^31
# run the loop from '0' to '31' only
for i in range(0,32):
# right shift both the numbers by 'i' and
# check if the bit at the 0th position is different
if ((( A >> i) & 1) != (( B >> i) & 1)):
count=count+1
print("Number of different bits :",count)
# Driver code
A = 12
B = 15
# find number of different bits
solve( A, B)
# This code is contributed by ihritik
C#
// C# implementation of the approach
using System;
class GFG
{
// compute number of different bits
static void solve(int A, int B)
{
int count = 0;
// since, the numbers are less than 2^31
// run the loop from '0' to '31' only
for (int i = 0; i < 32; i++) {
// right shift both the numbers by 'i' and
// check if the bit at the 0th position is different
if (((A >> i) & 1) != ((B >> i) & 1)) {
count++;
}
}
Console.WriteLine("Number of different bits : " + count);
}
// Driver code
public static void Main()
{
int A = 12, B = 15;
// find number of different bits
solve(A, B);
}
}
// This code is contributed by ihritik
PHP
> $i) & 1) != (($B >> $i) & 1)) {
$count++;
}
}
echo "Number of different bits : $count";
}
// Driver code
$A = 12;
$B = 15;
// find number of different bits
solve($A, $B);
// This code is contributed by ihritik
?>
Javascript
输出:
Number of different bits : 2