通过使用递归,我们可以将两个整数与给定的约束相乘。
若要将x和y相乘,请递归加xy次。
C++
// C++ program to Multiply two integers without
// using multiplication, division and bitwise
// operators, and no loops
#include
using namespace std;
class GFG
{
/* function to multiply two numbers x and y*/
public : int multiply(int x, int y)
{
/* 0 multiplied with anything gives 0 */
if(y == 0)
return 0;
/* Add x one by one */
if(y > 0 )
return (x + multiply(x, y-1));
/* the case where y is negative */
if(y < 0 )
return -multiply(x, -y);
}
};
// Driver code
int main()
{
GFG g;
cout << endl << g.multiply(5, -11);
getchar();
return 0;
}
// This code is contributed by SoM15242
C
#include
/* function to multiply two numbers x and y*/
int multiply(int x, int y)
{
/* 0 multiplied with anything gives 0 */
if(y == 0)
return 0;
/* Add x one by one */
if(y > 0 )
return (x + multiply(x, y-1));
/* the case where y is negative */
if(y < 0 )
return -multiply(x, -y);
}
int main()
{
printf("\n %d", multiply(5, -11));
getchar();
return 0;
}
Java
class GFG {
/* function to multiply two numbers x and y*/
static int multiply(int x, int y) {
/* 0 multiplied with anything gives 0 */
if (y == 0)
return 0;
/* Add x one by one */
if (y > 0)
return (x + multiply(x, y - 1));
/* the case where y is negative */
if (y < 0)
return -multiply(x, -y);
return -1;
}
// Driver code
public static void main(String[] args) {
System.out.print("\n" + multiply(5, -11));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Function to multiply two numbers
# x and y
def multiply(x,y):
# 0 multiplied with anything
# gives 0
if(y == 0):
return 0
# Add x one by one
if(y > 0 ):
return (x + multiply(x, y - 1))
# The case where y is negative
if(y < 0 ):
return -multiply(x, -y)
# Driver code
print(multiply(5, -11))
# This code is contributed by Anant Agarwal.
C#
// Multiply two integers without
// using multiplication, division
// and bitwise operators, and no
// loops
using System;
class GFG {
// function to multiply two numbers
// x and y
static int multiply(int x, int y) {
// 0 multiplied with anything gives 0
if (y == 0)
return 0;
// Add x one by one
if (y > 0)
return (x + multiply(x, y - 1));
// the case where y is negative
if (y < 0)
return -multiply(x, -y);
return -1;
}
// Driver code
public static void Main() {
Console.WriteLine(multiply(5, -11));
}
}
// This code is contributed by vt_m.
PHP
0 )
return ($x + multiply($x,
$y - 1));
/* the case where
y is negative */
if($y < 0 )
return -multiply($x, -$y);
}
// Driver Code
echo multiply(5, -11);
// This code is contributed by mits.
?>
Javascript
输出:
-55