Print Number Star Right Diamond Pattern Type3
PRINTING PATTERN:
4*4*4*4
3*3*3
2*2
1
1
2*2
3*3*3
4*4*4*4
PREREQUISITE:
Basic knowledge of C language and use of loops.
ALGORITHM:
- Take the number of rows as input from the user and store it in any variable.(‘r‘ in this case).
- Divide the value of ‘r’ by 2 and replace it in r. And give this value to count
- Run a loop ‘r’ number of times to iterate through each of the rows. From i=0 to i<r. The loop should be structured as for( i=0 ; i<r : i++).
- Run a loop from j=r to j>i. The loop should be structured as for(j=r; j>i ; j–)
- Run an if statement if(j!=r). If true the print star and count else only print count.
- Then in the outer if statement decrement count. Then print a newline
- Outside this loop increment count and run a loop from i=0 to i<r. The loop should be structured as for( i=0 ; i<r : i++).
- Run a nested loop from j=0 to j<=i . The loop should be structured as for( j=0 ; j<=i ; j++). Inside the loop run an if statement if(j!=0) then print star and digit else print only digit.
- Outside the loop increment count and print a newline.
CODE IN C:
#include<stdio.h> int main() { int i,j,r,count;//declaring integer variables i,j for loops , r for number of rows printf("Enter the number of rows/columns :\n");//asking user for the number of rows; scanf("%d",&r);//taking number of rows and saving in variable r r=r/2; count=r; for(i=0;i<r;i++) //loop for number of rows { for(j=r;j>i;j--)//loop to print digit in every column of a row { if(j!=r) { printf("*%d",count);//printing digit } else { printf("%d",count);//printing digit } } count--; printf("\n");//printing newline } count++; //intialising count =3 for(i=0;i<r;i++) //loop for number of rows { for(j=0;j<=i;j++) //loop to print digit in every column of a row { if(j!=0) { printf("*%d",count);//printing digit } else { printf("%d",count);//printing digit } } count++; //incrementing count printf("\n"); //printing newline } }
TAKING INPUT:DISPLAYING OUTPUT:
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#include
int main()
{
int n,i,j,k,l=4,m=4,o,p,q;
scanf(“%d”,&n);
// enter ‘8’ as input
if(n>5)
{
for(i=4;i>0;i–)
{
if(n>=5)
{
for(j=m;j>0;j–)
{
printf(“%d”,i);
if(j!=1)
{
printf(“*”);
}
}
m–;
printf(“\n”);
n–;
}
}
}
if(n<=4)
{
for(o=1;o<5;o++)
{
for(p=l;p<5;p++)
{
printf("%d",o);
if((5-p)!=1)
{
printf("*");
}
}
l–;
printf("\n");
}
}
}
n=int(input())
for i in range(n,0,-1):
for j in range(1,i*2):
if j%2==0:
print(“*”,end=””)
else:
print(i,end=””)
print()
for i in range(1,n+1):
for j in range(1,i*2):
if j%2==0:
print(“*”,end=””)
else:
print(i,end=””)
print()