Given two numbers x, y which denotes the number of set bits. Also given is a number C. The task is to print the number of ways in which we can form two numbers A and B such that A has x number of set bits and B has y number of set bits and A+B = C.
Examples:
Input: X = 1, Y = 1, C = 3 Output: 2 So two possible ways are (A = 2 and B = 1) and (A = 1 and B = 2) Input: X = 2, Y = 2, C = 20 Output: 3
Recursive Approach: The above problem can be solved using recursion. Calculate the total count of pairs recursively. There will have 4 possibilities. We start from the rightmost bit position.
- If the bit position at C is 1, then there are four possibilities to get 1 at that index.
- If the carry is 0, then we can use 1 bit out of X and 0 bit out of Y, or the vice-versa which generates no carry for next step.
- If the carry is 1, then we can use 1 set bit’s from each which generates a carry for the next step else we use no set bits from X and Y which generates no carry.
- If the bit position at C is 0, then there are four possibilities to get 0 at that bit position.
- If the carry is 1, then we can use 1 set bit out of X and 0 bit out of Y or the vice versa which generates a carry of 1 for the next step.
- If the carry is 0, then we can use 1 and 1 out of X and Y respectively, which generates a carry of 1 for next step. We can also use no set bits, which generates no carry for the next step.
Below is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Recursive function to generate // all combinations of bits int func( int third, int seta, int setb,
int carry, int number)
{ // find if C has no more set bits on left
int shift = (number >> third);
// if no set bits are left for C
// and there are no set bits for A and B
// and the carry is 0, then
// this combination is possible
if (shift == 0 and seta == 0 and setb == 0 and carry == 0)
return 1;
// if no set bits are left for C and
// requirement of set bits for A and B have exceeded
if (shift == 0 or seta < 0 or setb < 0)
return 0;
// Find if the bit is 1 or 0 at
// third index to the left
int mask = shift & 1;
int ans = 0;
// carry = 1 and bit set = 1
if ((mask) && carry) {
// since carry is 1, and we need 1 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
ans += func(third + 1, seta, setb, 0, number)
+ func(third + 1, seta - 1, setb - 1, 1, number);
}
// carry = 0 and bit set = 1
else if (mask && !carry) {
// since carry is 0, and we need 1 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
ans += func(third + 1, seta - 1, setb, 0, number)
+ func(third + 1, seta, setb - 1, 0, number);
}
// carry = 1 and bit set = 0
else if (!mask && carry) {
// since carry is 1, and we need 0 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
ans += func(third + 1, seta - 1, setb, 1, number)
+ func(third + 1, seta, setb - 1, 1, number);
}
// carry = 0 and bit set = 0
else if (!mask && !carry) {
// since carry is 0, and we need 0 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
ans += func(third + 1, seta, setb, 0, number)
+ func(third + 1, seta - 1, setb - 1, 1, number);
}
// Return ans
return ans;
} // Function to count ways int possibleSwaps( int a, int b, int c)
{ // Call func
return func(0, a, b, 0, c);
} // Driver Code int main()
{ int x = 2, y = 2, c = 20;
cout << possibleSwaps(x, y, c);
return 0;
} |
// Java implementation of above approach import java.util.*;
public class GFG{
// Recursive function to generate // all combinations of bits static int func( int third, int seta, int setb,
int carry, int number)
{ // find if C has no more set bits on left
int shift = (number >> third);
// if no set bits are left for C
// and there are no set bits for A and B
// and the carry is 0, then
// this combination is possible
if (shift == 0 && seta == 0 && setb == 0 && carry == 0 )
return 1 ;
// if no set bits are left for C and
// requirement of set bits for A and B have exceeded
if (shift == 0 || seta < 0 || setb < 0 )
return 0 ;
// Find if the bit is 1 or 0 at
// third index to the left
int mask = (shift & 1 );
int ans = 0 ;
// carry = 1 and bit set = 1
if ((mask) == 1 && carry == 1 ) {
// since carry is 1, and we need 1 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
ans += func(third + 1 , seta, setb, 0 , number)
+ func(third + 1 , seta - 1 , setb - 1 , 1 , number);
}
// carry = 0 and bit set = 1
else if (mask == 1 && carry == 0 ) {
// since carry is 0, and we need 1 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
ans += func(third + 1 , seta - 1 , setb, 0 , number)
+ func(third + 1 , seta, setb - 1 , 0 , number);
}
// carry = 1 and bit set = 0
else if (mask == 0 && carry == 1 ) {
// since carry is 1, and we need 0 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
ans += func(third + 1 , seta - 1 , setb, 1 , number)
+ func(third + 1 , seta, setb - 1 , 1 , number);
}
// carry = 0 and bit set = 0
else if (mask == 0 && carry == 0 ) {
// since carry is 0, and we need 0 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
ans += func(third + 1 , seta, setb, 0 , number)
+ func(third + 1 , seta - 1 , setb - 1 , 1 , number);
}
// Return ans
return ans;
} // Function to count ways static int possibleSwaps( int a, int b, int c)
{ // Call func
return func( 0 , a, b, 0 , c);
} // Driver Code public static void main(String args[])
{ int x = 2 , y = 2 , c = 20 ;
System.out.println(possibleSwaps(x, y, c));
} } // This code is contributed by Samim Hossain Mondal. |
# Python implementation of above approach # Recursive function to generate # all combinations of bits def func(third, seta, setb, carry, number):
# find if C has no more set bits on left
shift = (number >> third);
# if no set bits are left for C
# and there are no set bits for A and B
# and the carry is 0, then
# this combination is possible
if (shift = = 0 and seta = = 0 and setb = = 0 and carry = = 0 ):
return 1 ;
# if no set bits are left for C and
# requirement of set bits for A and B have exceeded
if (shift = = 0 or seta < 0 or setb < 0 ):
return 0 ;
# Find if the bit is 1 or 0 at
# third index to the left
mask = (shift & 1 );
ans = 0 ;
# carry = 1 and bit set = 1
if ((mask) = = 1 and carry = = 1 ):
# since carry is 1, and we need 1 at C's bit position
# we can use 0 and 0
# or 1 and 1 at A and B bit position
ans + = func(third + 1 , seta, setb, 0 , number) + func(third + 1 , seta - 1 , setb - 1 , 1 , number);
# carry = 0 and bit set = 1
elif (mask = = 1 and carry = = 0 ):
# since carry is 0, and we need 1 at C's bit position
# we can use 1 and 0
# or 0 and 1 at A and B bit position
ans + = func(third + 1 , seta - 1 , setb, 0 , number) + func(third + 1 , seta, setb - 1 , 0 , number);
# carry = 1 and bit set = 0
elif (mask = = 0 and carry = = 1 ):
# since carry is 1, and we need 0 at C's bit position
# we can use 1 and 0
# or 0 and 1 at A and B bit position
ans + = func(third + 1 , seta - 1 , setb, 1 , number) + func(third + 1 , seta, setb - 1 , 1 , number);
# carry = 0 and bit set = 0
elif (mask = = 0 and carry = = 0 ):
# since carry is 0, and we need 0 at C's bit position
# we can use 0 and 0
# or 1 and 1 at A and B bit position
ans + = func(third + 1 , seta, setb, 0 , number) + func(third + 1 , seta - 1 , setb - 1 , 1 , number);
# Return ans
return ans;
# Function to count ways def possibleSwaps(a, b, c):
# Call func
return func( 0 , a, b, 0 , c);
# Driver Code if __name__ = = '__main__' :
x = 2 ;
y = 2 ;
c = 20 ;
print (possibleSwaps(x, y, c));
# This code is contributed by Rajput-Ji |
// C# implementation of above approach using System;
public class GFG {
// Recursive function to generate
// all combinations of bits
static int func( int third, int seta, int setb,
int carry, int number)
{
// find if C has no more set bits on left
int shift = (number >> third);
// if no set bits are left for C
// and there are no set bits for A and B
// and the carry is 0, then
// this combination is possible
if (shift == 0 && seta == 0 && setb == 0 && carry == 0)
return 1;
// if no set bits are left for C and
// requirement of set bits for A and B have exceeded
if (shift == 0 || seta < 0 || setb < 0)
return 0;
// Find if the bit is 1 or 0 at
// third index to the left
int mask = (shift & 1);
int ans = 0;
// carry = 1 and bit set = 1
if ((mask) == 1 && carry == 1) {
// since carry is 1, and we need 1 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
ans += func(third + 1, seta, setb, 0, number) +
func(third + 1, seta - 1, setb - 1, 1, number);
}
// carry = 0 and bit set = 1
else if (mask == 1 && carry == 0) {
// since carry is 0, and we need 1 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
ans += func(third + 1, seta - 1, setb, 0, number) +
func(third + 1, seta, setb - 1, 0, number);
}
// carry = 1 and bit set = 0
else if (mask == 0 && carry == 1) {
// since carry is 1, and we need 0 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
ans += func(third + 1, seta - 1, setb, 1, number) +
func(third + 1, seta, setb - 1, 1, number);
}
// carry = 0 and bit set = 0
else if (mask == 0 && carry == 0) {
// since carry is 0, and we need 0 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
ans += func(third + 1, seta, setb, 0, number) +
func(third + 1, seta - 1, setb - 1, 1, number);
}
// Return ans
return ans;
}
// Function to count ways
static int possibleSwaps( int a, int b, int c) {
// Call func
return func(0, a, b, 0, c);
}
// Driver Code
public static void Main(String []args) {
int x = 2, y = 2, c = 20;
Console.WriteLine(possibleSwaps(x, y, c));
}
} // This code is contributed by umadevi9616 |
<script> // javascript implementation of above approach // Recursive function to generate
// all combinations of bits
function func(third , seta , setb , carry , number) {
// find if C has no more set bits on left
var shift = (number >> third);
// if no set bits are left for C
// and there are no set bits for A and B
// and the carry is 0, then
// this combination is possible
if (shift == 0 && seta == 0 && setb == 0 && carry == 0)
return 1;
// if no set bits are left for C and
// requirement of set bits for A and B have exceeded
if (shift == 0 || seta < 0 || setb < 0)
return 0;
// Find if the bit is 1 or 0 at
// third index to the left
var mask = (shift & 1);
var ans = 0;
// carry = 1 and bit set = 1
if ((mask) == 1 && carry == 1) {
// since carry is 1, and we need 1 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
ans += func(third + 1, seta, setb, 0, number) +
func(third + 1, seta - 1, setb - 1, 1, number);
}
// carry = 0 and bit set = 1
else if (mask == 1 && carry == 0) {
// since carry is 0, and we need 1 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
ans += func(third + 1, seta - 1, setb, 0, number) +
func(third + 1, seta, setb - 1, 0, number);
}
// carry = 1 and bit set = 0
else if (mask == 0 && carry == 1) {
// since carry is 1, and we need 0 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
ans += func(third + 1, seta - 1, setb, 1, number) +
func(third + 1, seta, setb - 1, 1, number);
}
// carry = 0 and bit set = 0
else if (mask == 0 && carry == 0) {
// since carry is 0, and we need 0 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
ans += func(third + 1, seta, setb, 0, number) +
func(third + 1, seta - 1, setb - 1, 1, number);
}
// Return ans
return ans;
}
// Function to count ways
function possibleSwaps(a , b , c)
{
// Call func
return func(0, a, b, 0, c);
}
// Driver Code
var x = 2, y = 2, c = 20;
document.write(possibleSwaps(x, y, c));
// This code is contributed by Rajput-Ji </script> |
3
Efficient Approach: The above problem can be solved using bitmask DP.
- Initialize a 4-D DP array of size 64 * 64 * 64 * 2 as 10^18 has a maximum of 64 set bits with -1
- The first state of the DP array stores the number of bits traversed in C from right. The second state stores the number of set-bits used out of X and third state stores the number of set bits used out of Y. The fourth state is the carry bit which refers to the carry generated when we perform an addition operation.
- The recurrence will have 4 possibilities. We start from the rightmost bit position.
- If the bit position at C is 1, then there are four possibilities to get 1 at that index.
- If the carry is 0, then we can use 1 bit out of X and 0 bit out of Y, or the vice-versa which generates no carry for next step.
- If the carry is 1, then we can use 1 set bit’s from each which generates a carry for the next step else we use no set bits from X and Y which generates no carry.
- If the bit position at C is 0, then there are four possibilities to get 0 at that bit position.
- If the carry is 1, then we can use 1 set bit out of X and 0 bit out of Y or the vice versa which generates a carry of 1 for the next step.
- If the carry is 0, then we can use 1 and 1 out of X and Y respectively, which generates a carry of 1 for next step. We can also use no set bits, which generates no carry for the next step.
- Summation of all possibilities is stored in dp[third][seta][setb][carry] to avoid re-visiting same states.
Below is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Initial DP array int dp[64][64][64][2];
// Recursive function to generate // all combinations of bits int func( int third, int seta, int setb,
int carry, int number)
{ // if the state has already been visited
if (dp[third][seta][setb][carry] != -1)
return dp[third][seta][setb][carry];
// find if C has no more set bits on left
int shift = (number >> third);
// if no set bits are left for C
// and there are no set bits for A and B
// and the carry is 0, then
// this combination is possible
if (shift == 0 and seta == 0 and setb == 0 and carry == 0)
return 1;
// if no set bits are left for C and
// requirement of set bits for A and B have exceeded
if (shift == 0 or seta < 0 or setb < 0)
return 0;
// Find if the bit is 1 or 0 at
// third index to the left
int mask = shift & 1;
dp[third][seta][setb][carry] = 0;
// carry = 1 and bit set = 1
if ((mask) && carry) {
// since carry is 1, and we need 1 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
dp[third][seta][setb][carry]
+= func(third + 1, seta, setb, 0, number)
+ func(third + 1, seta - 1, setb - 1, 1, number);
}
// carry = 0 and bit set = 1
else if (mask && !carry) {
// since carry is 0, and we need 1 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
dp[third][seta][setb][carry]
+= func(third + 1, seta - 1, setb, 0, number)
+ func(third + 1, seta, setb - 1, 0, number);
}
// carry = 1 and bit set = 0
else if (!mask && carry) {
// since carry is 1, and we need 0 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
dp[third][seta][setb][carry]
+= func(third + 1, seta - 1, setb, 1, number)
+ func(third + 1, seta, setb - 1, 1, number);
}
// carry = 0 and bit set = 0
else if (!mask && !carry) {
// since carry is 0, and we need 0 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
dp[third][seta][setb][carry]
+= func(third + 1, seta, setb, 0, number)
+ func(third + 1, seta - 1, setb - 1, 1, number);
}
return dp[third][seta][setb][carry];
} // Function to count ways int possibleSwaps( int a, int b, int c)
{ memset (dp, -1, sizeof (dp));
// function call that returns the
// answer
int ans = func(0, a, b, 0, c);
return ans;
} // Driver Code int main()
{ int x = 2, y = 2, c = 20;
cout << possibleSwaps(x, y, c);
return 0;
} |
// Java implementation of the above approach import java.util.*;
class GfG {
// Initial DP array static int dp[][][][] = new int [ 64 ][ 64 ][ 64 ][ 2 ];
// Recursive function to generate // all combinations of bits static int func( int third, int seta, int setb,
int carry, int number)
{ // if the state has already been visited
if (dp[third][seta][setb][carry] != - 1 )
return dp[third][seta][setb][carry];
// find if C has no more set bits on left
int shift = (number >> third);
// if no set bits are left for C
// and there are no set bits for A and B
// and the carry is 0, then
// this combination is possible
if (shift == 0 && seta == 0 && setb == 0 && carry == 0 )
return 1 ;
// if no set bits are left for C and
// requirement of set bits for A and B have exceeded
if (shift == 0 || seta < 0 || setb < 0 )
return 0 ;
// Find if the bit is 1 or 0 at
// third index to the left
int mask = shift & 1 ;
dp[third][seta][setb][carry] = 0 ;
// carry = 1 and bit set = 1
if ((mask == 1 ) && carry == 1 ) {
// since carry is 1, and we need 1 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
dp[third][seta][setb][carry]
+= func(third + 1 , seta, setb, 0 , number)
+ func(third + 1 , seta - 1 , setb - 1 , 1 , number);
}
// carry = 0 and bit set = 1
else if (mask == 1 && carry == 0 ) {
// since carry is 0, and we need 1 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
dp[third][seta][setb][carry]
+= func(third + 1 , seta - 1 , setb, 0 , number)
+ func(third + 1 , seta, setb - 1 , 0 , number);
}
// carry = 1 and bit set = 0
else if (mask == 0 && carry == 1 ) {
// since carry is 1, and we need 0 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
dp[third][seta][setb][carry] += func(third + 1 , seta - 1 , setb, 1 , number)
+ func(third + 1 , seta, setb - 1 , 1 , number);
}
// carry = 0 and bit set = 0
else if (mask == 0 && carry == 0 ) {
// since carry is 0, and we need 0 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
dp[third][seta][setb][carry] += func(third + 1 , seta, setb, 0 , number)
+ func(third + 1 , seta - 1 , setb - 1 , 1 , number);
}
return dp[third][seta][setb][carry];
} // Function to count ways static int possibleSwaps( int a, int b, int c)
{ for ( int q = 0 ; q < 64 ; q++)
{
for ( int r = 0 ; r < 64 ; r++)
{
for ( int p = 0 ; p < 64 ; p++)
{
for ( int d = 0 ; d < 2 ; d++)
{
dp[q][r][p][d] = - 1 ;
}
}
}
}
// function call that returns the
// answer
int ans = func( 0 , a, b, 0 , c);
return ans;
} // Driver Code public static void main(String[] args)
{ int x = 2 , y = 2 , c = 20 ;
System.out.println(possibleSwaps(x, y, c));
} } |
# Python3 implementation of the above approach # Initial DP array dp = [[[[ - 1 , - 1 ] for i in range ( 64 )]
for j in range ( 64 )]
for k in range ( 64 )]
# Recursive function to generate # all combinations of bits def func(third, seta, setb, carry, number):
# if the state has already been visited
if dp[third][seta][setb][carry] ! = - 1 :
return dp[third][seta][setb][carry]
# find if C has no more set bits on left
shift = number >> third
# if no set bits are left for C
# and there are no set bits for A and B
# and the carry is 0, then
# this combination is possible
if (shift = = 0 and seta = = 0 and
setb = = 0 and carry = = 0 ):
return 1
# if no set bits are left for C and
# requirement of set bits for A and B have exceeded
if (shift = = 0 or seta < 0 or setb < 0 ):
return 0
# Find if the bit is 1 or 0 at
# third index to the left
mask = shift & 1
dp[third][seta][setb][carry] = 0
# carry = 1 and bit set = 1
if (mask) and carry:
# since carry is 1, and we need 1 at
# C's bit position we can use 0 and 0
# or 1 and 1 at A and B bit position
dp[third][seta][setb][carry] + = \
func(third + 1 , seta, setb, 0 , number) + \
func(third + 1 , seta - 1 , setb - 1 , 1 , number)
# carry = 0 and bit set = 1
elif mask and not carry:
# since carry is 0, and we need 1 at C's
# bit position we can use 1 and 0
# or 0 and 1 at A and B bit position
dp[third][seta][setb][carry] + = \
func(third + 1 , seta - 1 , setb, 0 , number) + \
func(third + 1 , seta, setb - 1 , 0 , number)
# carry = 1 and bit set = 0
elif not mask and carry:
# since carry is 1, and we need 0 at C's
# bit position we can use 1 and 0
# or 0 and 1 at A and B bit position
dp[third][seta][setb][carry] + = \
func(third + 1 , seta - 1 , setb, 1 , number) + \
func(third + 1 , seta, setb - 1 , 1 , number)
# carry = 0 and bit set = 0
elif not mask and not carry:
# since carry is 0, and we need 0 at C's
# bit position we can use 0 and 0
# or 1 and 1 at A and B bit position
dp[third][seta][setb][carry] + = \
func(third + 1 , seta, setb, 0 , number) + \
func(third + 1 , seta - 1 , setb - 1 , 1 , number)
return dp[third][seta][setb][carry]
# Function to count ways def possibleSwaps(a, b, c):
# function call that returns the answer
ans = func( 0 , a, b, 0 , c)
return ans
# Driver Code if __name__ = = "__main__" :
x, y, c = 2 , 2 , 20
print (possibleSwaps(x, y, c))
# This code is contributed by Rituraj Jain |
// C# implementation of the above approach using System;
class GFG
{ // Initial DP array static int [,,,]dp = new int [64, 64, 64, 2];
// Recursive function to generate // all combinations of bits static int func( int third, int seta, int setb,
int carry, int number)
{ // if the state has already been visited
if (third > -1 && seta > -1 &&
setb > -1 && carry > -1)
if (dp[third, seta, setb, carry] != -1)
return dp[third, seta, setb, carry];
// find if C has no more set bits on left
int shift = (number >> third);
// if no set bits are left for C
// and there are no set bits for A and B
// and the carry is 0, then
// this combination is possible
if (shift == 0 && seta == 0 &&
setb == 0 && carry == 0)
return 1;
// if no set bits are left for C and
// requirement of set bits for A and
// B have exceeded
if (shift == 0 || seta < 0 || setb < 0)
return 0;
// Find if the bit is 1 or 0 at
// third index to the left
int mask = shift & 1;
dp[third, seta, setb, carry] = 0;
// carry = 1 and bit set = 1
if ((mask == 1) && carry == 1)
{
// since carry is 1, and we need 1 at
// C's bit position, we can use 0 and 0
// or 1 and 1 at A and B bit position
dp[third, seta,
setb, carry] += func(third + 1, seta,
setb, 0, number) +
func(third + 1, seta - 1,
setb - 1, 1, number);
}
// carry = 0 and bit set = 1
else if (mask == 1 && carry == 0)
{
// since carry is 0, and we need 1 at
// C's bit position, we can use 1 and 0
// or 0 and 1 at A and B bit position
dp[third, seta,
setb, carry] += func(third + 1, seta - 1,
setb, 0, number) +
func(third + 1, seta,
setb - 1, 0, number);
}
// carry = 1 and bit set = 0
else if (mask == 0 && carry == 1)
{
// since carry is 1, and we need 0 at
// C's bit position, we can use 1 and 0
// or 0 and 1 at A and B bit position
dp[third, seta,
setb, carry] += func(third + 1, seta - 1,
setb, 1, number) +
func(third + 1, seta,
setb - 1, 1, number);
}
// carry = 0 and bit set = 0
else if (mask == 0 && carry == 0)
{
// since carry is 0, and we need 0 at
// C's bit position, we can use 0 and 0
// or 1 and 1 at A and B bit position
dp[third, seta,
setb, carry] += func(third + 1, seta,
setb, 0, number) +
func(third + 1, seta - 1,
setb - 1, 1, number);
}
return dp[third, seta, setb, carry];
} // Function to count ways static int possibleSwaps( int a, int b, int c)
{ for ( int q = 0; q < 64; q++)
{
for ( int r = 0; r < 64; r++)
{
for ( int p = 0; p < 64; p++)
{
for ( int d = 0; d < 2; d++)
{
dp[q, r, p, d] = -1;
}
}
}
}
// function call that returns the
// answer
int ans = func(0, a, b, 0, c);
return ans;
} // Driver Code public static void Main(String[] args)
{ int x = 2, y = 2, c = 20;
Console.WriteLine(possibleSwaps(x, y, c));
} } // This code is contributed by Rajput-Ji |
<script> // javascript implementation of above approach // Recursive function to generate
// all combinations of bits
function func(third , seta , setb , carry , number) {
// find if C has no more set bits on left
var shift = (number >> third);
// if no set bits are left for C
// and there are no set bits for A and B
// and the carry is 0, then
// this combination is possible
if (shift == 0 && seta == 0 && setb == 0 && carry == 0)
return 1;
// if no set bits are left for C and
// requirement of set bits for A and B have exceeded
if (shift == 0 || seta < 0 || setb < 0)
return 0;
// Find if the bit is 1 or 0 at
// third index to the left
var mask = (shift & 1);
var ans = 0;
// carry = 1 and bit set = 1
if ((mask) == 1 && carry == 1) {
// since carry is 1, and we need 1 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
ans += func(third + 1, seta, setb, 0, number) +
func(third + 1, seta - 1, setb - 1, 1, number);
}
// carry = 0 and bit set = 1
else if (mask == 1 && carry == 0) {
// since carry is 0, and we need 1 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
ans += func(third + 1, seta - 1, setb, 0, number) +
func(third + 1, seta, setb - 1, 0, number);
}
// carry = 1 and bit set = 0
else if (mask == 0 && carry == 1) {
// since carry is 1, and we need 0 at C's bit position
// we can use 1 and 0
// or 0 and 1 at A and B bit position
ans += func(third + 1, seta - 1, setb, 1, number) +
func(third + 1, seta, setb - 1, 1, number);
}
// carry = 0 and bit set = 0
else if (mask == 0 && carry == 0) {
// since carry is 0, and we need 0 at C's bit position
// we can use 0 and 0
// or 1 and 1 at A and B bit position
ans += func(third + 1, seta, setb, 0, number) +
func(third + 1, seta - 1, setb - 1, 1, number);
}
// Return ans
return ans;
}
// Function to count ways
function possibleSwaps(a , b , c)
{
// Call func
return func(0, a, b, 0, c);
}
// Driver Code
var x = 2, y = 2, c = 20;
document.write(possibleSwaps(x, y, c));
// This code contributed by Rajput-Ji </script> |
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