please tell me how to design in C++, which generate sequence of numbers uptil 20 entries using function?

Question

please tell me how to design a program in C++, which generate sequence of numbers uptil 20 entries using function??

Answer 1

Here is an easier program for you.



#include



using namespace std;



void sequence(int number);



int main()

{

    int number; // For the sequence amount

    cout << "Enter the number for the senquence: ";

    cin >> number;

    // You might want to check if the number is negative or over 20.

    sequence(number);

}



void sequence(int number)

{

    for (int i = 0; i <= 20; i = i + number)

    {

        cout << i << " ";

    }

}



Hope that helps. You may want i to start at 1 and not 0, or even start at number.

for (int i = number; i < 20; i = i + number)

Answer 2

numbers are tobe in what sequence??

I can tell u roughly-

define class

make function(void)

call ti from diferent function

for counting uptil 20 use a loop

Answer 3

EDITED:



#include

#include

using namespace std;



const int SIZE(20);

const int MAX_RANGE(100);



// FUNCTION generates sequence of random numbers

void generate(int array_type[], const int array_size, const int max_range)

{

int index(0);



// Initialize the random generator

srand( time(NULL) );



// Generates 'array_size' numbers between 0 and MAX_RANGE

while(index!=array_size)

array_type[index++] = rand()%max_range;

}



// FUNCTION generates sequence of sequential numbers

void generate(int array_type[], const int array_size, const int start, const int step)

{

int index(0);



// assign value of 'start' to the first index

// then increment 'index' by 1

array_type[index++] = start;



// generates sequence of sequential numbers

while(index!=array_size)

array_type[index++] = start + (step*index);

}



int main()

{

// array to holds the generated numbers

int sequence[SIZE];



// call to function generate

generate(sequence, SIZE, 0, 2);



// Output the sequence

int index(0);

while(index!=SIZE)

cout<
cout<


return EXIT_SUCCESS;

}