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C program 001
For n = 5, the pattern should be as follows,
XXXXXXXXX
XX - XXX -XX
X - X - X -X -X
XX -XXX - XX
XXXXXXXXX
XX - XXX- XX
X- X- X - X - X
XX - XXX -XX
XXXXXXXXX
Hint: Number of rows = 2*n - 1
Write a program to print the pattern for the given ‘n’ value.
Input format:
Input is an integer which corresponds to the n.
Output format:
Refer the sample output.
Sample Intput 1:
5
Sample Output 1:
XXXXXXXXX
XX - XXX -XX
X - X - X -X -X
XX -XXX - XX
XXXXXXXXX
XX - XXX- XX
X- X- X - X - X
XX - XXX -XX
XXXXXXXXX
Sample Input 2:
9
Sample Output 2:
XXXXXXXXXXXXXXXXX
XX-----XXX-----XX
X-X---X-X-X---X-X
X--X-X--X--X-X--X
X---X---X---X---X
X--X-X--X--X-X--X
X-X---X-X-X---X-X
XX-----XXX-----XX
XXXXXXXXXXXXXXXXX
XX-----XXX-----XX
X-X---X-X-X---X-X
X--X-X--X--X-X--X
X---X---X---X---X
X--X-X--X--X-X--X
X-X---X-X-X---X-X
XX-----XXX-----XX
XXXXXXXXXXXXXXXXX
Solution Code :
#include<stdio.h>
int main()
{
int n;
int r;
scanf("%d",&n);
r=2*n-1;
int a[r][r],l=r-1;
printf("\n");
int q=r/2-1,p=r/2+1,e=1,d=l-1;
for(int i=1;i<r;i++)
{
if(i<r/2)
{
a[i][q]='X';
q--;
a[i][p]='X';
p++;
}
if(i>r/2)
{
a[i][e]='X';
e++;
a[i][d]='X';
d--;
}
}
for(int i=0;i<r;i++)
{
for(int j=0;j<r;j++)
{
if(i==j||j==l||j==r-1||j==0||i==0||i==r-1||j==r/2||i==r/2)
a[i][j]='X';
else if(a[i][j]=='X')
{
printf("%c",a[i][j]);
continue;
}
else
a[i][j]='-';
printf("%c",a[i][j]);
}
l--;
printf("\n");
}
}
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