Internal Varsity Number Square Printing
PRINTING PATTERN:
333
313
323
333
PREREQUISITE:
Basic knowledge of C language and loops.
ALGORITHM:
- Take number of rows as input from the user and save it in any variable (‘r’in this case).
- Run a loop ‘r’ number of times to iterate through the rows. From i=0 to i<r. The loop should be structured as for( i=0 ; i<r ; i++).
- Inside this if condition use an id condition to print the top and bottom rows. if(i==0 || i==r-1).
- Inside this run a nested loop to iterate through the columns. From j=0 to j<3. The loop should be structured as for(j=0 ; j<3 ; j++).
- Print “3” in this loop.
- Write an else condition. Inside it a run a similar loop . From j=0 to j<3. The loop should be structured as for(j=0 ; j<3 ; j++).
- Use an if condition. if(j==0 || j==2). And print “3”.
- Else print the iterating variable for rows.
- Inside the main loop print a newline
CODE IN C:
#include<stdio.h> int main() { int i,j,r; //declaring integer variables i,j for loops , r for number of rows printf("Enter the number of rows :\n"); //asking user for input scanf("%d",&r); //saving number of rows in variable r for(i=0;i<r;i++) //loop for number of rows { if(i==0 || i==r-1) //if condition to print first and last row { for(j=0;j<3;j++) //loop to print first and last row { printf("3"); //printing 3 } } else //else to print rest of the square { for(j=0;j<3;j++) //loop to print the rest of the square { if(j==0 || j==2) //if condition to print the outer values { printf("3"); //printing 3 } else//else condition to { printf("%d",i); //Printing the middle column } } } printf("\n"); //printing newline } }
TAKING INPUT:
DISPLAYING OUTPUT:
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Simplest code in c++ convert it in any other language.
oid pattern3(int n){
for(int i=0; i<n; ++i){
for(int j=0; j<3; ++j){
if(i==0 || i==n-1 ||j==0||j==2){
cout<<"3";
}else{
cout<<i;
}
}
cout<<endl;
}
}
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C++ Solution
#include
using namespace std;
int main()
{
int n;
cin>>n;
for(int i=1;i<=n;i++)//Loop
{
if(i==1||i==n)//condition checking
{
for(int j=1;j<n;j++)//Loop
cout<<"3";
}
else
{
for(int j=1;j<n;j++)//Loop
{
if(j==1||j==3)
cout<<"3";
else
cout<<i-1;
}
}
cout<<endl;
}
return 0;
}
Python 3
n=int(input())
for i in range(n):
for j in range(n):
if(i==0 or j==0 or j==n-1 or i==n-1 or j!=n//2):
print(n,end=”)
else:
print(i,end=”)
print(‘\n’)
n=int(input())
if n%2!=0 and n>2:
for i in range(1,n+2):
for j in range(0,n):
if(i==1 or i==n+1 or j==0 or j==n-1 or j>n//2 or j<n//2):
print(n,end='')
else:
print(i-1,end='')
print()