Practice programming questions loops
(SAMPLE SOLUTIONS AT BOTTOM OF PAGE)

  1. Write a program that inputs a positive integer from the user and then displays all the positive integers that are factors of that number.
    Hint: X is a factor of Y if and only if (Y % X) is zero.

  2. Write a function that computes N! for positive integers (i.e. 1*2*...*N) then use that function to write a program that can read two positive integers from the user, M and N, and compute N choose M.
    Note: N choose M is defined as N! / (M! * (N-M)!)

  3. Modify the program below so that it only computes the sum of elements that are on the diagonals. 
    #include <cstdio>

const int SIZE = 24;

int main()
{
   int sum = 0;
   int position = 0;
   for (int row = 0; row < SIZE; row++) {
       for (int column = 0; column < SIZE; column++) {
           printf("%d ", position);
           sum = sum + position;
           position++;
       }
       printf("\n");
   }
   printf( "The sum is %d\n", sum);
}

    #include <iostream>
using namespace std;

const int SIZE = 24;

int main()
{
   int sum = 0;
   int position = 0;
   for (int row = 0; row < SIZE; row++) {
       for (int column = 0; column < SIZE; column++) {
           cout << position;
           sum = sum + position;
           position++;
       }
       cout << endl;
   }
   cout << "The sum is " << sum << endl;
}
SAMPLE SOLUTION

  1. Factors of N 
    #include <cstdio>

int main()
{
   // get the positive integer from the user
   //    (no error checking done here)
   printf( "Please enter a positive integer\n");
   int N;
   scanf("%d", &N);

   // for each value 1..N see if it evenly divides N
   //     i.e. remainder of 0 
   for (int n = 1; n <= N; n++) {
       if ((N % n) == 0) {
          printf("%d is a factor of %d\n", n, N);
       }
   }
   return 0;
}

  2. Factorials and N choose M 
    int factorial(int N) 
{
   // result will be the product 1*2*3*...*N
   int result = 1;
   for (int n = 1; n <= N; n++) {
       result *= n;
   }
   return result;
}

void choose()
{
   // get N and M from the user
   printf( "Please enter two positive integers\n");
   int N, M;
   scanf("%d", &N);
   scanf("%d", &M);

   // result = N choose M
   int result = 0;
   if (N > M) {
      result = factorial(N) / (factorial(M) * factorial(N-M));
   } 
   printf("%d choose %d is %d\n", N, M, result);
}

  3. Modify the prog
    Only one change needed, shown in italics, to determine if the current position is on the forward or backward diagonal 
    #include <cstdio>

const int SIZE = 24;

int main()
{
   int sum = 0;
   int position = 0;
   for (int row = 0; row < SIZE; row++) {
       for (int column = 0; column < SIZE; column++) {
           printf("%d ", position); 
           if ((row == column) || ((row+column) == (SIZE-1))) 
               sum = sum + position;
           position++;
       }
       printf("\n");
   }
   printf("The sum is %d\n", sum);
}