C++ pattern output - c++

Question

I am working on a project that requires me to have a loop output a pattern into the console.
I have to use a for loop in my code. I've gotten to a point where I can only get half of the pattern onto the screen but the rest does not appear in the console.
My code:
#include <iostream>
using namespace std;
int main()
{
int i, j;
for (i=5; i>=1; i--)
{
for (j=1; j != i; j++)
{
cout << "5";
cout << "#";
cout << endl;
}
}
return 0;
}
So this code outputs:
####5
###5
##5
#5
5
But I need it to output:
####5
###5#
##5##
#5###
5####
How would I change my code to get it to show that output?
I hope that makes sense,
Thank you

Answer 1

your internal loop is the one that tracks moving 5 from right to left
for (i=5; i >= 1; i--) {
for (j=1; j <= 5; j++) {
cout << (( i == j ) ? "5" : "#");
}
cout << endl;
}
removing the fancy things
for (i=5; i >= 1; i--)
{
for (j=1; j <= 5; j++)
{
if( i == j )
cout << "5";
else
cout << "#";
}
cout << endl;
}

Answer 2

This is what you are looking for.
#include <iostream>
using namespace std;
int main()
{
int i,j;
for (i=4; i>=0; i--)
{
for(int z=0; z<5;z++){
if(z==i){
cout <<"5";
}
else{
cout <<"#";
}
}
cout <<"\n";
}
return 0;
}
####5
###5#
##5##
#5###
5####
Chears :-)

Answer 3

Let's be clear: the OP's code is completely wrong and I cannot understand how it works and produces that output, so I wrote it from scratch.
Before_edit:
I can't understand how your code work, so I rewrite that from scratch.
#include <iostream>
void output_sharp(int cnt)
{
while (cnt != 0)
std::cout << "#";
}
int main(int argc, char *argv[])
{
for (int i = 0, j = 4; i != 5; ++i, --j) {
output_sharp(j - i);
std::cout << "5";
output_sharp(4 - j);
}
return 0;
}