给定三个正整数X ， Y和Z ，任务是计算在X和Y中需要翻转的最小位数，以使X OR Y(X | Y)等于Z。
例子：
Input : X = 5 Y = 8 Z = 6
Output : 3
Explanation :
Before After
Flipping Flipping
X :: 0101 0100
Y :: 1000 0010
------ -----
Z :: 0110 0110
So we need to flip 3 bits in order
to make (X | Y) = Z .
Input : X = 1 Y = 2 Z = 3
Output : 0方法 ：
为了解决此问题，我们需要观察到仅在以下情况下才需要翻转X或Y：
- 如果设置了Z中的当前位，而未同时设置X和Y中的相应位，则需要在X或Y中设置一个位。
- 如果未设置Z中的当前位，则需要取消设置A和B中的相应位(以设置为准)。如果两者都设置，则同时取消设置。
因此，我们需要遍历X，Y和Z的位，并按照上述两个步骤来获得所需的结果。
下面是上述方法的实现：
C++
// C++ Program to minimize
// number of bits to be fipped
// in X or Y such that their
// bitwise or is equal to minimize
#include
using namespace std;
// This function returns minimum
// number of bits to be flipped in
// X and Y to make X | Y = Z
int minimumFlips(int X, int Y, int Z)
{
int res = 0;
while (X > 0 || Y > 0 || Z > 0) {
// If current bit in Z is set and is
// also set in either of X or Y or both
if (((X & 1) || (Y & 1)) && (Z & 1)) {
X = X >> 1;
Y = Y >> 1;
Z = Z >> 1;
continue;
}
// If current bit in Z is set
// and is unset in both X and Y
else if (!(X & 1) && !(Y & 1) && (Z & 1)) {
// Set that bit in either X or Y
res++;
}
else if ((X & 1) || (Y & 1) == 1)
{
// If current bit in Z is unset
// and is set in both X and Y
if ((X & 1) && (Y & 1) && !(Z & 1)) {
// Unset the bit in both X and Y
res += 2;
}
// If current bit in Z is unset and
// is set in either X or Y
else if (((X & 1) || (Y & 1)) && !(Z & 1))
{
// Unset that set bit
res++;
}
}
X = X >> 1;
Y = Y >> 1;
Z = Z >> 1;
}
return res;
}
// Driver Code
int main()
{
int X = 5, Y = 8, Z = 6;
cout << minimumFlips(X, Y, Z);
return 0;
} Java
// Java program to minimize
// number of bits to be fipped
// in X or Y such that their
// bitwise or is equal to minimize
import java.util.*;
class GFG{
// This function returns minimum
// number of bits to be flipped in
// X and Y to make X | Y = Z
static int minimumFlips(int X, int Y, int Z)
{
int res = 0;
while (X > 0 || Y > 0 || Z > 0)
{
// If current bit in Z is set and is
// also set in either of X or Y or both
if (((X % 2 == 1) ||
(Y % 2 == 1)) &&
(Z % 2 == 1))
{
X = X >> 1;
Y = Y >> 1;
Z = Z >> 1;
continue;
}
// If current bit in Z is set
// and is unset in both X and Y
else if (!(X % 2 == 1) &&
!(Y % 2 == 1) &&
(Z % 2 == 1))
{
// Set that bit in either X or Y
res++;
}
else if ((X % 2 == 1) || (Y % 2 == 1))
{
// If current bit in Z is unset
// and is set in both X and Y
if ((X % 2 == 1) &&
(Y % 2 == 1) &&
!(Z % 2 == 1))
{
// Unset the bit in both X and Y
res += 2;
}
// If current bit in Z is unset and
// is set in either X or Y
else if (((X % 2 == 1) ||
(Y % 2 == 1)) &&
!(Z % 2 == 1))
{
// Unset that set bit
res++;
}
}
X = X >> 1;
Y = Y >> 1;
Z = Z >> 1;
}
return res;
}
// Driver Code
public static void main(String[] args)
{
int X = 5, Y = 8, Z = 6;
System.out.print(minimumFlips(X, Y, Z));
}
}
// This code is contributed by amal kumar choubeyPython3
# Python 3 Program to minimize
# number of bits to be fipped
# in X or Y such that their
# bitwise or is equal to minimize
# This function returns minimum
# number of bits to be flipped in
# X and Y to make X | Y = Z
def minimumFlips(X, Y, Z):
res = 0
while (X > 0 or Y > 0 or
Z > 0):
# If current bit in Z is
# set and is also set in
# either of X or Y or both
if (((X & 1) or (Y & 1)) and
(Z & 1)):
X = X >> 1
Y = Y >> 1
Z = Z >> 1
continue
# If current bit in Z is set
# and is unset in both X and Y
elif (not (X & 1) and not(Y & 1) and
(Z & 1)):
# Set that bit in either
# X or Y
res += 1
elif ((X & 1) or (Y & 1) == 1):
# If current bit in Z is unset
# and is set in both X and Y
if ((X & 1) and (Y & 1) and
not(Z & 1)):
# Unset the bit in both
# X and Y
res += 2
# If current bit in Z is
# unset and is set in
# either X or Y
elif (((X & 1) or (Y & 1)) and
not(Z & 1)):
# Unset that set bit
res += 1
X = X >> 1
Y = Y >> 1
Z = Z >> 1
return res
# Driver Code
if __name__ == "__main__":
X = 5
Y = 8
Z = 6
print(minimumFlips(X, Y, Z))
# This code is contributed by ChitranayalC#
// C# program to minimize
// number of bits to be fipped
// in X or Y such that their
// bitwise or is equal to minimize
using System;
class GFG{
// This function returns minimum
// number of bits to be flipped in
// X and Y to make X | Y = Z
static int minimumFlips(int X, int Y, int Z)
{
int res = 0;
while (X > 0 || Y > 0 || Z > 0)
{
// If current bit in Z is set and is
// also set in either of X or Y or both
if (((X % 2 == 1) ||
(Y % 2 == 1)) &&
(Z % 2 == 1))
{
X = X >> 1;
Y = Y >> 1;
Z = Z >> 1;
continue;
}
// If current bit in Z is set
// and is unset in both X and Y
else if (!(X % 2 == 1) &&
!(Y % 2 == 1) &&
(Z % 2 == 1))
{
// Set that bit in either X or Y
res++;
}
else if ((X % 2 == 1) || (Y % 2 == 1))
{
// If current bit in Z is unset
// and is set in both X and Y
if ((X % 2 == 1) &&
(Y % 2 == 1) &&
!(Z % 2 == 1))
{
// Unset the bit in both X and Y
res += 2;
}
// If current bit in Z is unset and
// is set in either X or Y
else if (((X % 2 == 1) ||
(Y % 2 == 1)) &&
!(Z % 2 == 1))
{
// Unset that set bit
res++;
}
}
X = X >> 1;
Y = Y >> 1;
Z = Z >> 1;
}
return res;
}
// Driver Code
public static void Main()
{
int X = 5, Y = 8, Z = 6;
Console.Write(minimumFlips(X, Y, Z));
}
}
// This code is contributed by Code_MechJavascript
输出：
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