给定两个数字中可能的特殊对的数量
给定两个数字 A，B，任务是找出 A，B 的特殊对的数字，两个数字 A，B 的特殊对是满足两个给定条件的一对数字 X，Y–A = X | Y，B = X & Y. 示例:
Input: A = 3, B = 0
Output: 2
(0, 3), (1, 2) will satisfy the conditions
Input: A = 5, B = 7
Output: 0
方法:这里的关键观察是，如果我们希望两个数字 X，Y 的 or 等于 A，那么 X，Y 都必须小于或等于 A，如果有人大于 A，那么 OR 就不会等于 A，这将给出我们的循环将终止的极限，剩下的我们将尝试检查两对是否满足给定的条件，然后我们将递增计数器。 以下是所需的实现:
C++
// C++ implementation of above approach
#include <iostream>
using namespace std;
// Function to count the pairs
int countPairs(int A, int B)
{
// Variable to store a number of special pairs
int cnt = 0;
for (int i = 0; i <= A; ++i) {
for (int j = i; j <= A; ++j) {
// Calculating AND of i, j
int AND = i & j;
// Calculating OR of i, j
int OR = i | j;
// If the conditions are met,
// then increment the count of special pairs
if (OR == A and AND == B) {
cnt++;
}
}
}
return cnt;
}
// Driver code
int main()
{
int A = 3, B = 0;
cout << countPairs(A, B);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of above approach
class GFG
{
// Function to count the pairs
static int countPairs(int A, int B)
{
// Variable to store a number
// of special pairs
int cnt = 0;
for (int i = 0; i <= A; ++i)
{
for (int j = i; j <= A; ++j)
{
// Calculating AND of i, j
int AND = i & j;
// Calculating OR of i, j
int OR = i | j;
// If the conditions are met,
// then increment the count
// of special pairs
if (OR == A && AND == B)
{
cnt++;
}
}
}
return cnt;
}
// Driver code
public static void main(String [] args)
{
int A = 3, B = 0;
System.out.println(countPairs(A, B));
}
}
// This code is contributed by ihritik
Python 3
# Python3 implementation of above
# approach
# Function to count the pairs
def countPairs(A,B):
# Variable to store a number
# of special pairs
cnt=0
for i in range(0,A+1):
for j in range(i,A+1):
# Calculating AND of i, j
AND = i&j
OR = i|j
# If the conditions are met,
# then increment the count of
# special pairs
if(OR==A and AND==B):
cnt +=1
return cnt
if __name__=='__main__':
A = 3
B = 0
print(countPairs(A,B))
# This code is contributed by
# Shrikant13
C
// C# implementation of above approach
using System;
class GFG
{
// Function to count the pairs
static int countPairs(int A, int B)
{
// Variable to store a number
// of special pairs
int cnt = 0;
for (int i = 0; i <= A; ++i)
{
for (int j = i; j <= A; ++j)
{
// Calculating AND of i, j
int AND = i & j;
// Calculating OR of i, j
int OR = i | j;
// If the conditions are met,
// then increment the count
// of special pairs
if (OR == A && AND == B)
{
cnt++;
}
}
}
return cnt;
}
// Driver code
public static void Main()
{
int A = 3, B = 0;
Console.WriteLine(countPairs(A, B));
}
}
// This code is contributed by ihritik
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of above approach
// Function to count the pairs
function countPairs($A, $B)
{
// Variable to store a number
// of special pairs
$cnt = 0;
for ($i = 0; $i <= $A; ++$i)
{
for ($j = $i; $j <= $A; ++$j)
{
// Calculating AND of i, j
$AND = $i & $j;
// Calculating OR of i, j
$OR = $i | $j;
// If the conditions are met,
// then increment the count
// of special pairs
if ($OR == $A && $AND == $B)
{
$cnt++;
}
}
}
return $cnt;
}
// Driver code
$A = 3;
$B = 0;
echo countPairs($A, $B);
// This code is contributed by ihritik
?>
java 描述语言
<script>
// Javascript implementation of above approach
// Function to count the pairs
function countPairs(A, B) {
// Variable to store a number of special pairs
let cnt = 0;
for (let i = 0; i <= A; ++i) {
for (let j = i; j <= A; ++j) {
// Calculating AND of i, j
let AND = i & j;
// Calculating OR of i, j
let OR = i | j;
// If the conditions are met,
// then increment the count of special pairs
if (OR == A && AND == B) {
cnt++;
}
}
}
return cnt;
}
// Driver code
let A = 3, B = 0;
document.write(countPairs(A, B));
// This code is contributed by _saurabh_jaiswal
<script>
Output:
2
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