Write a program to print the above pattern
#include <stdio.h>
int main()
{
int r,c, n,nr,nc,c1=1,c2=1;
scanf("%d", &n);
nr=2*n-1;
nc=2*n-1;
for(r=1;r<=nr;r++)
{
for(c=1;c<=nc;c++)
{
if(c<c1||c>c2)
printf(" ");
else
printf("*");
}
if(r<n)
{
c1++;
c2=c2+2;
}
else
{
c1--;
c2=c2-2;
}
printf("\n");
}
}
==================================================
http://chitranshugupta.blogspot.com/p/c.html#pt6
/* Write a Program to print Butterfly Matrix Pattern
* *
** **
*** ***
**** ****
*********
**** ****
*** ***
** **
* *
*/
int main()
{
int n=5;//No. of rows
int i=1;
int c1=1;
int c2=2*n-1;
int j=n+1;
int b1=n;
int b2=n;
for(int row=1;row<=(2*n-1);row++)
{
for(int col=1;col<=(2*n-1);col++)
{
if((row==i && col >c1 && col< c2) || (row==j && col >=b1 && col<=b2))
printf(" ");
else
printf("*");
}
if(row< n)
{
i++;
c1++;
c2--;
}
else if(row >n)
{
j++;
b1--;
b2++;
}
printf("\n");
}
return 0;
}
Write a C program to print the following number pattern
#include <stdio.h>
main()
{
int n,i,j,k,spaces,prints,c;
scanf("%d",&n);
spaces = n-1;
prints = 1;
for(i=1;i<=2*n-1;i++)
{
for(j=1;j<=spaces;j++)
printf(" ");
c=1;
for(k=1;k<= 2*prints -1; k++)
{
printf("%3d",c);
if(c<n)
c++;
else
c--;
}
if(i<n)
{
spaces--;
prints++;
}
else
{
spaces++;
prints--;
}
printf("\n");
}
}
====================================================================
Write a C program to print the following pattern in C
#include <stdio.h>
main()
{
int n,i,j,k,l;
scanf("%d",&n);
int n1=2*n-1;
int t=n1/2,c=0;
for(i=1;i<=n;i++)
{
for(j=1;j<=t;j++)
printf("*");
for(k=1;k<=c;k++)
printf(" ");
for(;j<=(n1-c);j++)
printf("*");
if(c == 0)
c=c+1;
else
{
c=c+2;
t=t-1;
}
printf("\n");
}
}
===================================================================
Write a C program to print the following pattern
Program:
#include <stdio.h>
int max(int a,int b)
{
if(a>b)
return a;
return b;
}
main()
{
int n,i,j,l;
scanf("%d",&n);
l=2*n+1;
for(i=0;i<l;i++){
for(j=0;j<l;j++)
printf("%d",max(abs(n-i),abs(n-j)));
printf("\n");
}
}
Write a C program to print the following pattern
1
1 2
4 3 9
#include<stdio.h>
main()
{
int n,c=1,flag=1,i,j,k;
scanf("%d",&n);
for(i=0;i<n;i++)
{
for(j=n-1;j>i;j--)
{
printf(" ");
}
for(k=0;k<=i;k++)
{
if(flag){
printf("%3d",c);flag=0;}
else{
printf("%3d",c*c);flag=1;
c++;
}
}
printf("\n");
}
}
========================================================================
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
in the
above pattern for example n= 1211
1 2 1 1
units
and tens place digits are same digit ‘1’ and count =2 ,j =1, n1 =0
so n1=
(10*count+same_digit)*j +n1
n1=(10*2+1)*1+0=21
and n
becomes 12 i.e, n =12
in 12,
units and tens place digit are not same. digit 2 occurs once so count =1 ,
j=100 , n1=21
n1 = (10*1+2)*100+21
= 1221
now n
=1
1
occurs only once so count = 1, j= 10000,n1=1221
n1=(10*1+1)*10000+1221
n1 =
11*10000+1221= 111221
#include
<stdio.h>
int
main()
{
int i,n=1,j,n1,temp,c,rows;
printf("enter number of
rows\n");
scanf("%d",&rows);
for(i=0;i<rows;i++)
{
n1=0;
j=1;
printf("%d\n",n);
while(n)
{
temp=n%10;
n=n/10;
c=1;
while(temp
== n%10) // number of digits same
starting from units place
{
c++;
n=n/10;
}
n1=(((10*c)+temp)*j)+n1;
j=j*100;
}
n=n1;
}
return 0;
}
============================================
to print
*
* *
* * *
in first line, we print one * and
in second line, we print * * (2 stars) and
in third line, we print * * * (3 stars) and
as i is giving the line number,so j runs i times.
to print i *'s in ith line
========================================
n=4
* * * *
* * * *
* * * *
* * * *
in first line, we print * * * * (n stars) and in second line, we print * * * * (n stars) and in third line, we print * * * * (n stars) and so j runs n times. to print n *'s in each line
