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C Program to print different type patterns

1
1 2
1 2 3
1 2 3 4

 #include<stdio.h>
#include<conio.h> 
void main()
{
   int n=0, i=0, j=0;
   printf("enter the limit");
   scanf("%d",&n);
   
   for(i=1; i<=n; i++)
     {
       for(j=1; j<=i; j++)
         {
            printf("%d ",j); 
          }
         printf("\n");
      }
}

Output of the above program is as following.

4 3 2 1
4 3 2
4 3
4

 #include<stdio.h> 
#include<conio.h>
void main()
{
   int n=0, i=0, j=0;
   printf("enter the limit");
   scanf("%d",&n);
       
   for(i=1; i<=n; i++)
     {
       for(j=n; j>=i; j--)
          {
             printf("%d ", j); 
          }
        printf("\n");
     }
}

Output of the above program is as following.

&nbsp&nbsp &nbsp &nbsp &nbsp *
&nbsp &nbsp &nbsp * *
&nbsp &nbsp* * *
* * * *

 #include<stdio.h>
#include<conio.h> 
void main()
{
   int n=0, i=0, j=0;
   printf("enter the limit");
   scanf("%d",&n);
       
   for(i=1; i<=n; i++)
     {
     //this loop is used to print spaces 
         for(j=n; j>i; j--)
            {
               printf(" "); 
            }
         
     //this loop used to print the stars
         for(j=1;j<=i;j++)
            {
               printf("* "); 
            }
         printf("\n");
      }
}
 

Output of the above program is as following.

1 1
2 2 3 3
4 4 5
4 4 3 3
2 2
1 1

#include<stdio.h>
#include<conio.h> 
int main()
{
int i=1,limit=0,j=0,row=1,k=1,f=0,flag=0;	
printf("enter the limit");
scanf("%d",&limit);

for(i=1; i<=((2*limit)-1); i++)
{
    k=1;
    flag=0;
   	
  for(j=1;j<=((2*limit)-1); j++)
  {    
     if(k==row)
     {
      printf("%d",k);
     }
     else
      {
       printf(" ");
      }
     if(flag==0)
     {
        k++;
     }
     else
     {
       k--;
     }
      if(k==limit)
     {
       flag=1;	
     }
	
   }	
if(f==0)
{
row++;
}
else
{
row--;
}
if(row==limit)
{
f=1;	
}	
printf("\n");	
}
	
}

Output of the above program is as following.

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