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[NEWBIE] C++ Diamond ( * ) problem using loops
ZwischenzugHello, I have solved a problem about Diamond asterisks using loops. However, my code is too long and is there any solution that can provide a better and much shorter code?
My code looks like this. Thank you :)#include <iostream>
int main()
{
using namespace std;
int x, y, n, m;
for (x = 1; x <= 5; x++)
{
n = x % 2; //determines if even or odd
switch (x)
{
case 1:
for (y = 1; y <= 5; y++)
{
m = y % 2; //determines if even or odd
if ((m == 1) && (y != 3))
{
cout<<" ";
}
else if (m == 0)
{
cout<<" ";
}
else if ((m == 1) && (y = 3))
{
cout<<"*";
}
}
cout<<"\n";
break;
case 2:
for (y = 1; y <= 5; y++)
{
m = y % 2; //determines if even or odd
if ((m == 1) && (y != 3))
{
cout<<" ";
}
else if (m == 0)
{
cout<<"*";
}
else if ((m == 1) && (y = 3))
{
cout<<"*";
}
}
cout<<"\n";
break;
case 3:
for (y = 1; y <= 5; y++)
{
cout<<"*";
}
cout<<"\n";
break;
case 4:
for (y = 1; y <= 5; y++)
{
m = y % 2; //determines if even or odd
if ((m == 1) && (y != 3))
{
cout<<" ";
}
else if (m == 0)
{
cout<<"*";
}
else if ((m == 1) && (y = 3))
{
cout<<"*";
}
}
cout<<"\n";
break;
case 5:
for (y = 1; y <= 5; y++)
{
m = y % 2; //determines if even or odd
if ((m == 1) && (y != 3))
{
cout<<" ";
}
else if (m == 0)
{
cout<<" ";
}
else if ((m == 1) && (y = 3))
{
cout<<"*";
}
}
cout<<"\n";
break;
}
}
return 0;
}
