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## Multiplication, Subtraction, Division of Integers Using Summation Operator

In this task, we can only use the summation operator. In such tasks, it is useful to recall the essence of mathematical operations and their implementation by adding (or using other operations).

## Subtraction

How do you implement the subtraction using summation? It’s very simple. Operation x – y is the same thing as x + (-1) * y. Since we can not use the multiplication operator, we have to create a negation function.

[code language=”cpp”]

public static int negation(int x) {

int neg = 0;

int y = x < 0 ? 1 : -1;

while (x != 0) {

neg += y;

x += y;

}

return neg;

}

public static int subtract(int x, int y) {

return x + negate(y);

}

[/code]

A negative value of k is obtained by summing the number -1, k times.

## Multiplication

The relationship between summation and multiplication is also quite obvious. To multiply x and y, we need to add x value to itself, y times.

[code language=”cpp”]

public static int mult(int x, int y) {

if (x < y) { return mult(y, x); } int sum = 0; for (int i = mod(y); i > 0; i–) {

sum += x;

}

if (y < 0) {

sum = negation(sum);

}

return sum;

}

public static int mod(int x) {

if (x < 0) {

return negation(x);

} else {

return x;

}

}

[/code]

We need to pay special attention to the negative numbers. If x – y is a negative number, it is necessary to take into account the sign of the sum of:

[code language=”cpp”]

multipl (x, y) <- abs (y) * x * (-1 if y <0).

[/code]

In addition, to solve this problem, we created a simple function mod.

## Division

The most difficult mathematical operation is division. It is a good idea to use the divide multiplication, subtraction and negation methods for the implementation of the division.

We need to find k, if k = x / y. Let us restate the problem: find k when x = yk. Now we change the terms so that the problem could be solved with the help of the already known operation – multiplication.

Note that k can be calculated as the result of the summation of b, until a is obtained. The number of copies of y, required to obtain x, is the value of k.

Of course, this decision cannot be called a full-fledged division, but it works. But using such implementation makes finding the remainder of the division impossible.

The following code implements this algorithm:

[code language=”cpp”]

public int division(int x, int y)

throws java.lang.ArithmeticException {

if ( y == 0) {

throw new java.lang.ArithmeticException("ERROR");

}

int moda = mod(x);

int modb = mod(y);

int result = 0;

int k = 0;

while (result + modb <= moda) {

result += modb;

k++;

}

if ((x < 0 && y < 0) || (x > 0 && y > 0)) {

return k;

} else {

return negation(k);

}

}

[/code]

Thanks for your attention!