要求翻转的最小位数，以使A和B的按位或等于C

鉴于3个阳性整数A，B和C，该任务是在计数A和B所需要的位，使得位或A和B等于C或不翻转的的最小数目。

例子：

Input: A = 2, B = 2, C = 3
Output: 1
Explanation:
The binary representation of A is 010, B is 010 and C is 011. 
Flip the 3rd bit of either A or B, such that A | B = C, i.e. 011 | 010 = 011.
Therefore, the total number of flips required is 1.

Input: A = 2, B = 6, C = 5
Output: 3

方法：请按照以下步骤解决问题：

• 初始化一个变量，例如res ，该变量存储所需的最少翻转位数。
• 遍历A ， B和C的每一位，然后执行以下步骤：
  • 如果C的^第i^个位没有设置，然后检查以下内容：
    • 如果A的^第i^个位被设置，增量RES与1，I A的^第i位需要被翻转。
    • 如果B的^第i^个位被设置，增量RES与1，I B的^第i位需要被翻转。
  • 如果C的^第i位设置，然后检查以下内容：
    • 如果未同时设置A和B的^第i位，则将res增大1 ，则任一位都需要翻转。
    • 如果同时设置了A和B的^第i位，则无需翻转任何位。
• 完成上述步骤后，打印res的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count the number of
// bit flips required on A and B
// such that Bitwise OR of A and B is C
int minimumFlips(int A, int B, int C)
{
    // Stores the count of flipped bit
    int res = 0;
 
    // Iterate over the range [0, 32]
    for (int i = 0; i < 32; i++) {
 
        int x = 0, y = 0, z = 0;
 
        // Check if i-th bit of A is set
        if (A & (1 << i)) {
            x = 1;
        }
 
        // Check if i-th bit of B is
        // set or not
        if (B & (1 << i)) {
            y = 1;
        }
 
        // Check if i-th bit of C is
        // set or not
        if (C & (1 << i)) {
            z = 1;
        }
 
        // If i-th bit of C is unset
        if (z == 0) {
 
            // Check if i-th bit of
            // A is set or not
            if (x) {
                res++;
            }
 
            // Check if i-th bit of
            // B is set or not
            if (y) {
                res++;
            }
        }
 
        // Check if i-th bit of C
        // is set or not
        if (z == 1) {
 
            // Check if i-th bit of
            // both A and B is set
            if (x == 0 && y == 0) {
                res++;
            }
        }
    }
 
    // Return the count
    // of bits flipped
    return res;
}
 
// Driver Code
int main()
{
    int A = 2, B = 6, C = 5;
    cout << minimumFlips(A, B, C);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to count the number of
// bit flips required on A and B
// such that Bitwise OR of A and B is C
static int minimumFlips(int A, int B, int C)
{
     
    // Stores the count of flipped bit
    int res = 0;
 
    // Iterate over the range [0, 32]
    for(int i = 0; i < 32; i++)
    {
        int x = 0, y = 0, z = 0;
 
        // Check if i-th bit of A is set
        if ((A & (1 << i)) != 0)
        {
            x = 1;
        }
 
        // Check if i-th bit of B is
        // set or not
        if ((B & (1 << i)) != 0)
        {
            y = 1;
        }
 
        // Check if i-th bit of C is
        // set or not
        if ((C & (1 << i)) != 0)
        {
            z = 1;
        }
 
        // If i-th bit of C is unset
        if (z == 0)
        {
 
            // Check if i-th bit of
            // A is set or not
            if (x == 1)
            {
                res++;
            }
 
            // Check if i-th bit of
            // B is set or not
            if (y == 1)
            {
                res++;
            }
        }
 
        // Check if i-th bit of C
        // is set or not
        if (z == 1)
        {
 
            // Check if i-th bit of
            // both A and B is set
            if (x == 0 && y == 0)
            {
                res++;
            }
        }
    }
 
    // Return the count
    // of bits flipped
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
    int A = 2, B = 6, C = 5;
     
    System.out.println(minimumFlips(A, B, C));
}
}
 
// This code is contributed by Kingash

输出：

3

时间复杂度： O(1)
辅助空间： O(1)