Classroom practice: Basic loops

···o
··ooo
·ooooo
ooooooo

The principle of the solution

To solve the task it is recommended to draw a sketch to a sheet of paper. Having counted the number of leading spaces and the number of letter "o" in the subsequent lines, the solution follows.

• First try to render row i. Observe that there are (n-i) leading spaces and 2*i-1 letter "o" in that row.
• The leading spaces and letter "o"-s should be printed by loops.
• An other loop is needed to iterate over the lines to render.

#include <stdio.h>

int main(void) {
    int height;
    printf("Enter the height of the triangle: ");
    scanf("%d", &height);

    /* loop for the subsequent lines */
    for (int row = 1; row <= height; row = row+1) {
        /* printing the leading space characters */
        for (int x = 1; x <= height-row; x = x+1)
            printf(" ");
        /* printing the letter "o"-s */
        for (int x = 1; x <= row*2-1; x = x+1)
            printf("o");
        /* end of the current row, go to the next line */
        printf("\n");
    }

    return 0;
}

The principle of the solution

• We have to go on trying the divisions till number 1 is obtained. Hence, we will need a loop like while (number != 1)
• If the number is divisible with the current divisor candidate, we perform the division and print it. If it is not divisible, an other candidate divisor is chosen.
• When trying to find divisors, we always start with the smallest one and move toward the bigger ones.
• Is it a problem if the candidate divisor is not a prime number (like 4, or 6)? Why?

In the C programming language the % operator denotes the remainder of the division (dividend % divisor → remainder, for instance 9%7 yields 2). This is the easiest way to check divisibility.

The solution below always prints an entire row at once, for instance 150|2. Printing this row is necessary when we have the check for divisibility already done, and the remainder turned out to be 0. The output 150|2 means that 150 is divided by 2; hence this line must be printed before actually performing the division, to be able to print the number before the division. Consequently, the last line, consisting of number 1, must be printed separately at the end.

#include <stdio.h>

int main(void) {
    int number;
    printf("Enter a number: ");
    scanf("%d", &number);

    int divisor;
    divisor = 2;
    while (number > 1) {
        if (number % divisor == 0) { /* if divisible */
            printf("%5d|%d\n", number, divisor);
            number = number / divisor;
        }
        else
            divisor = divisor + 1;  /* not divisable */
    }
    printf("%5d\n", 1);

    return 0;
}

Somebody might prefer an alternative solution, where the number to be divided is printed first, and the divisor is printed only afterwards, after passing the test for divisibility. This is a correct solution as well. In this case the number (as given by the user) must be printed before the loop. At the same time, the "1" does not have to be printed at the end, as it will appear at the last division.

#include <stdio.h>

int main(void) {
    int number;
    printf("Enter a number: ");
    scanf("%d", &number);

    int divisor;
    divisor = 2;
    printf("%5d|", number);
    while (number > 1) {
        if (number % divisor == 0) { /* if divisible */
            printf("%d\n", divisor);
            number = number / divisor;
            printf("%5d|", number);
        }
        else
            divisor = divisor + 1; /* not divisible */
    }

    return 0;
}

It might be worth considering to pre-compute the prime numbers beforehands and use them as candidate divisors in the algorithm (instead of incrementing the candidate divisor by one every time). How much would the program be more efficient?

The principle of the solution

• During the conversion, we need to go from the bigger roman numbers to the smaller ones.
• What we have already printed to the screen, can be substracted from the variable. For instance, in case of 89, after having 80 printed (LXXX) only 9 remains to convert, which can be converted as if the user input was 9.

Let us draw a flow-chart or a structogram before writing the C program! Before formalizing the algorithm try testing it with a couple of examples, to explore the order of printing of the subsequent roman characters. Observe that, while XC (90), L (50) etc. can occur only once, some other values (including X (10)) can occur several times!

#include <stdio.h>

int main(void) {
    int x;

    printf("Enter a number: ");
    scanf("%d", &x);
    if (x < 1)
        printf("The number must be positive.\n");
    else if (x > 99)
        printf("Only numbers less than 100 are supported.\n");
    else {
        if (x >= 90) { printf("XC"); x = x-90; }
        if (x >= 50) { printf("L"); x = x-50; }
        if (x >= 40) { printf("XL"); x = x-40; }
        while (x >= 10) { printf("X"); x = x-10; }
        if (x >= 9) { printf("IX"); x = x-9; }
        if (x >= 5) { printf("V"); x = x-5; }
        if (x >= 4) { printf("IV"); x = x-4; }
        while (x >= 1) { printf("I"); x = x-1; }
        printf("\n");
    }

    return 0;
}