给定两个正整数A和B使得A!= B ,任务是找到一个使表达式(A AND X)*(B AND X)最大化的正整数X。
例子:
Input: A = 9 B = 8
Output: 8
(9 AND 8) * (8 AND 8) = 8 * 8 = 64 (maximum possible)
Input: A = 11 and B = 13
Output: 9
天真的方法:可以运行从1到max(A,B)的循环,并且可以轻松地找到使给定表达式最大化的X。
高效的方法:众所周知,
(a – b)2 ≥ 0
which implies (a + b)2 – 4*a*b ≥ 0
which implies a * b ≤ (a + b)2 / 4
Hence, it concludes that a * b will be maximum when a * b = (a + b)2 / 4
which implies a = b
From the above result, (A AND X) * (B AND X) will be maximum when (A AND X) = (B AND X)
现在X可以找到为:
A = 11 = 1011
B = 13 = 1101
X = ? = abcd
At 0th place: (1 AND d) = (1 AND d) implies d = 0, 1 but to maximize (A AND X) * (B AND X) d = 1
At 1st place: (1 AND d) = (0 AND d) implies c = 0
At 2nd place: (0 AND d) = (1 AND d) implies b = 0
At 3rd place: (1 AND d) = (1 AND d) implies a = 0, 1 but to maximize (A AND X) * (B AND X) a = 1
Hence, X = 1001 = 9
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
#define MAX 32
// Function to find X according
// to the given conditions
int findX(int A, int B)
{
int X = 0;
// int can have 32 bits
for (int bit = 0; bit < MAX; bit++) {
// Temporary ith bit
int tempBit = 1 << bit;
// Compute ith bit of X according to
// given conditions
// Expression below is the direct
// conclusion from the illustration
// we had taken earlier
int bitOfX = A & B & tempBit;
// Add the ith bit of X to X
X += bitOfX;
}
return X;
}
// Driver code
int main()
{
int A = 11, B = 13;
cout << findX(A, B);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
static int MAX = 32;
// Function to find X according
// to the given conditions
static int findX(int A, int B)
{
int X = 0;
// int can have 32 bits
for (int bit = 0; bit < MAX; bit++)
{
// Temporary ith bit
int tempBit = 1 << bit;
// Compute ith bit of X according to
// given conditions
// Expression below is the direct
// conclusion from the illustration
// we had taken earlier
int bitOfX = A & B & tempBit;
// Add the ith bit of X to X
X += bitOfX;
}
return X;
}
// Driver code
public static void main(String []args)
{
int A = 11, B = 13;
System.out.println(findX(A, B));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the approach
MAX = 32
# Function to find X according
# to the given conditions
def findX(A, B) :
X = 0;
# int can have 32 bits
for bit in range(MAX) :
# Temporary ith bit
tempBit = 1 << bit;
# Compute ith bit of X according to
# given conditions
# Expression below is the direct
# conclusion from the illustration
# we had taken earlier
bitOfX = A & B & tempBit;
# Add the ith bit of X to X
X += bitOfX;
return X;
# Driver code
if __name__ == "__main__" :
A = 11; B = 13;
print(findX(A, B));
# This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
static int MAX = 32;
// Function to find X according
// to the given conditions
static int findX(int A, int B)
{
int X = 0;
// int can have 32 bits
for (int bit = 0; bit < MAX; bit++)
{
// Temporary ith bit
int tempBit = 1 << bit;
// Compute ith bit of X according to
// given conditions
// Expression below is the direct
// conclusion from the illustration
// we had taken earlier
int bitOfX = A & B & tempBit;
// Add the ith bit of X to X
X += bitOfX;
}
return X;
}
// Driver code
public static void Main(String []args)
{
int A = 11, B = 13;
Console.WriteLine(findX(A, B));
}
}
// This code is contributed by 29AjayKumar
Javascript
9
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