write a program to obtain the following pattern A D C B E F G H I P O N M L K J # # # # # # # # # J K L M N O P I H G F E B C D A

Answer 1

Just did this on the bus. output is right but obviously this is way too much code. this way, it'd be easier to just do this:

cout<<"A"<<endl<<"DCB"<<endl<<"EFGHI"<<endl<<"PONMLKJ"<<endl<<"#########"<<endl<<"JKLMNOP"<<endl<<"IHGFE"<<endl<<"BCD"<<endl<<"A";

The above will yield the desired output in one cout statement but that'd be cheating really.

Answer 2

#include <iostream>
using namespace std;

int main() {

int A = 64;
int x = 1;
int y = 2;
int z = 1;
bool b = true;
char ch;

while (x>=1)
{
if (x==9)
{
cout<<"#########"<<endl;
A+=1;
y = -2;
x += y;
b = !b;
z=-1;
}

else if (b)
{
int B = A + x * z;
while (A!=B)
{
A+=z;
ch=A;
cout<<ch;
}
cout<<endl;
x += y;
b = !b;
}

else if (!b)
{
int B = A;
A+=x*z;
while(B!=A)
{
ch=A;
cout<<ch;
A-=z;

}
cout<<endl;
A+=x*z;
x+=y;
b=!b;
}
}

return 0;
}

Answer 3

Comments part 4

Now b is false, so the second else if statement starts running. Here, we want to count backwards, so now we copy A into B
B = A;
then, we want to count DOWN x steps so we take A and add x
A += x;
and count back to B in another while loop:

while(A !=B) //as long as A and B are not equal
{
ch = A ; //get ASCII value of A
cout<<ch; // write ch x times
A -= 1 //A minus one
}

we have counted backwards, so after the loop, A is the same as before, so we add x back to A
A += x;
then,
x += y; //x + 2 or after x==9 x-2
b = !b; // set b back to true

and that's it really. just one more thing.
after x==9 we want to count backwards.
so whenever we did
B = A + x ... A + 1 ..... or
B =A ... A += x ... A - 1

in the loops, to count back to 65 after x==9, all + becomes - and all - becomes + like this:

B = A - x ... A - 1 ...
B = A + x ... A + 1

for this, we add one last variable z
int z = 1;

and set it to minus one in the if(x==9)
z = -1; // we could also write z *= -1;

now in the else if parts, we use z wherever we need + and - to change when we count back down.

B += x; //becomes
B +=x*z; //B += x*z is the same as
B = B + (x*z) , z is always 1 or -1, when z is one,
it doesn't do anything, x*1 is still x.
when counting back down, z is minus one so
x* -1 = -x , now B + (-x ) equals B - x , so
B+=x*z with z = -1 is the same as
B -= x

That should explain it all, in case anyone was wondering. I've been in a car accident recently and now I've way too much time on my hands so I'm more than happy to answer any questions

Answer 4

/************************
Comment section:

A 'char' really holds a number that corresponds to a character on the ASCII table. Here, we need cpiral Letters A - P , ASCII code for A is 65, B is 66 and so on. (# is 35 but I didn't use it in the code)
To output an A we can convert a number like this:

int Num = 65;
char ch;
ch = Num;
cout<<ch;

So now we need this:

65
68 67 66
69 70 71 72 73
80 79 78 77 76 75 74
# # # # # # # # #
74 75 76 77 78 79 80
73 72 71 70 69
66 67 68
65

Answer 5

Comments part 2

...so I need to start with a line of one (65), then three (68 67 66) then five characters, so starting with ONE, every new line has TWO more characters, we do this:

int x = 1; // starting with ONE
int y = 2; // in the code do x += y so its 2 more

but, every second row counts backwards (68 67 66)
AND after x == 9 (the line with nine characters) we want the lines to go back to 7, then 5 , 3 and finally one character. ALSO after the line with 9 characters, we count backwards to 65.

the line with x==9 is different too because in that line there are no Letters, just a '#', nine times.

so after we declared our variables (int A = 64; char ch; int x= 1; int y = 2, there will be more later) the first thing we do is check wether the line is the one with nine characters:

if(x==9)
{
//code...
}

what we do here is
cout<<"# # # # # # # # #"<<endl;
to write the nine '#' characters
Then,
y = -2;
remember we start with x=1 and then we add y=2 so the number of characters in every line is two more. now at x==9, y becomes -2. the same code that we used to count upwards (x += y) will now count back for us, because x + -2 is like x-2.

so now we get

1
3
5
7
9 //write ######### , y = -2 to count back
7
5
3
1
-1
-3
-5
...

...now the loop goes on forever, first x+y counts up in steps of two, then at x==9, y becomes -2 and we count down forever. so the first thing after declaring our variables should be

while(x>=1)
{
//all code goes here
}

this way, our code will run only as long as x is one or more than one, and as soon as x is negative the code stops.

then inside the while loop we have three parts:
checking if x==9 and two more things.

Answer 6

comments part 3

we've checked for (x==9). Now we need the remaining code. Every other line needs to be reversed. We can add a boolean variable
bool b = true;
b can only have two values, true and false.
so if we do this

if(x==0)
{//code}

else if(b)
{
//some code
b = !b;
}

else if(!b)
{
//some code
b = !b;
}

this way, we start with b = true, so the first else if statement with else if(b) is run. inside the first else if, we do b = !b. this means, if b is true, it becomes false, if b is false, it becomes true. so inside the first else if, b is set to false.
the second else if will run if b is false
else if (!b)
and inside that else if , b is set to true, so the two will run one after the other as long as x is one or greater, and the first if statement will run if x is 9.

In the first else if, we will do the normal counting, in the second, the backwards counting.

so in the else if (b)

We take x (the number of characters in the line)
then we make a temporary variable int B, that only exists inside the else if.
We want to count up from A, so we take A plus the number of steps we want to count (x) and put the result in B
int B = A + x;
then
while (A!=B) //while A and B are different
{
A +=1; //A plus one
ch = A; //turn A into an ASCII character
cout<<ch; //write ch
}
after the loop has run x times, we do
cout<<endl; //new line
x += y; // x plus two, or minus two after x==9
b = !b; change b from true to false

Answer 7

OVO

Answer 8

Andres pls ans my remaining questions too

Answer 9

1 is the answer