How do you write a C plus plus program to find the first 5 multiples of a number?

Answer 1

#include<iostream>

#include<vector>

std::vector<size_t> multiples(size_t const num, const size_t terms)

{

std::vector<size_t> result;

size_t term=0;

while (++term<=terms)

result.push_back (num*term);

return result;

}

int main()

{

const std::vector<size_t> mults = multiples (42, 5);

std::cout << "The first 5 multiples of 42 are:";

for (auto value : mults) std::cout << '\t' << value;

std::cout << std::endl;

}

Answer 2

To write a program to check whether a particular number is a multiple of 5, or any other number for that matter, you check to see if the first number, modulus the second number, is zero.

if (N % 5 == 0) { /* N is divisible by 5 */ }

If you don't have the modulus (%) operator, you can use the explicit expression

if (N - int (N / 5) * 5 == 0) { /* N is divisible by 5 */ }

Answer 3

There is no modulus operator in C. Presumably you meant modulus in the literal sense to sum the absolute values of 5 signed integers. To obtain the absolute value of a signed integer, simply invoke the C standard library abs() function declared in <math.h>:

#include<math.h>

int sum_abs (int* arr, int size) {

int sum = 0;

for (int i=0; i<size; ++i) sum += abs (arr[i]);

return sum;

}

If you actually meant the modulo operator (%), it is not clear how you intend to use this operator to calculate a sum given that the modulo operator is a division operator (it returns the remainder after division) but it's not clear how the five values should be divided let alone what is to be summed.

Answer 4

The "%" operator gives the remainder of a division, for example: b = a % 5;