Pattern Printing coding 10

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CODE NO:- 12

 Print the below pattern

#LOGIC
Take two variables and assign it with value (((n*(n+1))/2)+2)-(n-1)  and 0(zero)  here n means the no. of entered rows and (((n*(n+1))/2)+2)-(n-1) is the first element of the pattern.
ex: if n is 4 then  (((n*(n+1))/2)+2)-(n-1) results 9. 
Now copy the value  (((n*(n+1))/2)+2)-(n-1) in other variable and increment it by 1 in the inner loop whereas in the outer loop decrement it by value (n-i-1). Proceed in the same manner to print the pattern.

 #Program( In C):

 In  C language we use  ' // '  for Single line comment and /*---*/ for multi line comment

 #include<stdio.h>// header file for i/p and o/p
int main()
{
int n,i,k,a,b=0;
printf("\n enter the no.of rows:");
scanf("%d",&n);
a=(((n*(n+1))/2)+2)-(n-1);
for(i=0;i<n;i++)
{
           b=a;
           for(k=0;k<n-i;k++)
  {
printf("%d ",b++); // incrementing the value column wise
           }
           a=a-(n-i-1); // decrementing the value row wise
printf("\n");
}
return 0;
}

 We got this output by using DEV C++ (IDE)

Link: For other pattern printing problems

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