CS50 - Stuck with mario pyramid [closed] - c

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has been doing pset1 "Mario Pyramid" and got stuck with the part about print "Hashes", already tried different ways but... nothing. In specific the problem is where it defines the value of "hs". https://docs.cs50.net/problems/mario/more/mario.html Here is about the problem pset1
#include <stdio.h>
#include <cs50.h>
int main(void)
{
int h, s, hs, i;
do {
h = get_int("Height: ");
} while (h < 0 || h > 23);
for (i = 0; i < h; i++) {
/* ignore 1 */
if (i < 1) {
printf("");
} else {
for (s = (h - i); s > 1; s--) {
/* Spaces */
printf(" ");
}
for (hs = 2; hs < h; hs++) {
/* Hashes */
printf("#");
}
/* Jump Line */
printf("\n");
}
}
}
Output:
Height: 6
####
####
####
####
####
When its supposed to look like a half pyramid

Answer 1

Before you start coding, try to understand how to solve this problem with pencil
and paper.
Specially when you do a course on computer science, it's more important to
understand the underlying principle more than the programming itself. Very often
you will need to figure out the pattern. Once you've done that, translating it
into a program is much easier.
So let us try to figure out the pattern here.
This is the pyramid you want
# #
## ##
### ###
#### ####
What is the first thing that you can observe? We only have to look at the left
half of the pyramid, because the right half is the same only mirrored.
So the problem becomes easier:
# | #
## --> | ##
### | ###
#### |####
I added | to make it clear where each line begin. So for a pyramid of 4:
we need 4 lines
every line has the same length, 4 characters
the number of spaces decrease and the number of hashes increase from top to
bottom:
for the first line we need 4-1=3 spaces and 1 hash
for the second line we need 4-2=2 spaces and 2 hashes
for the third line we need 4-3=1 spaces and 3 hashes
for the fourth line we need 4-4=0 spaces and 4 hashes
Are you beginning to see the pattern here?
So in computing we start counting at 0. Beginners often make the mistake to
start counting by 1. So let's repeat these sentences from 0:
for the 0th line we need 4-1-0=3 spaces and 0+1=1 hashes
for the 1st line we need 4-1-1=2 spaces and 1+1=2 hashes
for the 2nd line we need 4-1-2=1 spaces and 2+1=3 hashes
for the 3rd line we need 4-1-3=0 spaces and 3+1=4 hashes
thus for a pyramid of size n:
for the 0th line we need n-1-0 spaces and 0+1=1 hashes
for the 1st line we need n-1-1 spaces and 1+1=2 hashes
· · ·
for the n-1th line we need n-1-(n-1) spaces and (n-1)+1 hashes
Do you see the pattern again?
for the i th line we need n - 1 - i spaces and i+1 hashes.
We have the pattern for the left half, the right half is the same, just
mirrored. So now that we know the numbers of spaces and hashes, writing a loop
is very easy, you almost only have to plug in the formulas:
#include <stdio.h>
int main(void)
{
int n = 10, i, j;
for(i = 0; i < n; ++i)
{
// left half
for(j = 0; j < n-1-i; ++j)
printf(" ");
for(j = 0; j < i + 1; ++j)
printf("#");
// 2 spaces in the middle
printf(" ");
// right half, we swapped the order
for(j = 0; j < i + 1; ++j)
printf("#");
for(j = 0; j < n-1-i; ++j)
printf(" ");
printf("\n");
}
return 0;
}
Note that the right half reverses the order of the printing hashes and spaces.
But trailing spaces at the right can be omitted entirely, so the last for loop
can be removed:
// this for loop before the printf("\n");
// can be removed, not needed.
for(j = 0; j < n-1-i; ++j)
printf(" ");
When solving these kind of problems, try using this method, pick a
pencil and paper and solve it by hand. Doing
this we've figured out the patter and translating it into code was a piece of
cake. Notice that we even saw that the problem could be "cut in half" because
of the symmetry. Now an exercise for you: print the same pyramid but upside down.