不使用乘法、除法和按位运算符乘法两个整数，不循环
原文:https://www . geeksforgeeks . org/乘法-二进制数-不使用乘法-除法-按位-运算符-不循环/
通过使用递归，我们可以用给定的约束条件乘以两个整数。 要将 x 和 y 相乘，递归加 x 和 y 次。
C++
// C++ program to Multiply two integers without
// using multiplication, division and bitwise
// operators, and no loops
#include<iostream>
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<stdio.h>
/* 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 语言(一种计算机语言，尤用于创建网站)
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.
Python 3
# 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.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// function to multiply
// two numbers x and y
function 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
echo multiply(5, -11);
// This code is contributed by mits.
?>
java 描述语言
<script>
// javascript program to Multiply two integers without
// using multiplication, division and bitwise
// operators, and no loops
/* function to multiply two numbers x and y*/
function 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
document.write( multiply(5, -11));
// This code is contributed by todaysgaurav
</script>
输出:
-55
时间复杂度:O(y)，其中 y 是函数乘法()的第二个参数。
辅助空间:O(y) 俄罗斯农民(使用按位运算符乘以两个数字) 如果您发现上面的代码/算法有任何不正确的地方，请写评论，或者找到更好的方法来解决同样的问题。
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