The most common pattern is
for n=5
we have output
1
3*2
4*5*6
10*9*8*7
11*12*13*14*15
The Solution is:
#include <iostream>
using namespace std;
int main()
{
int n=5,num;
num=1;
int l=1;
int k=num;
for(int i=1;i<=n;i++)
{
k=num-1;
for(int j=1;j<=num;j++)
{
if(j%2==0)
cout<<"*";
else
{
if(i%2==0)
{
cout<<k+l-num+i;
l++;
k=k-2;
}
else
cout<<l++;
}
}
num=num+2;//the loop is going on as a prime number 1,3,5,7
cout<<"\n";
}
return 0;
}
for n=5
we have output
1
3*2
4*5*6
10*9*8*7
11*12*13*14*15
The Solution is:
#include <iostream>
using namespace std;
int main()
{
int n=5,num;
num=1;
int l=1;
int k=num;
for(int i=1;i<=n;i++)
{
k=num-1;
for(int j=1;j<=num;j++)
{
if(j%2==0)
cout<<"*";
else
{
if(i%2==0)
{
cout<<k+l-num+i;
l++;
k=k-2;
}
else
cout<<l++;
}
}
num=num+2;//the loop is going on as a prime number 1,3,5,7
cout<<"\n";
}
return 0;
}
Can u pls type the code in c
ReplyDeleteThis comment has been removed by the author.
Delete#include
Deleteint main() {
int n=5,k=1;
long l;
for(int i=1;i<=n;i++)
{
if(i%2!=0)
{
for(int j=1;j<=2*i-1;j++)
{
if(i%2!=0&&j%2==0)
{
printf("*");
}
else
{
printf("%d",k);
k++;
}
}printf("\n");
}
else
{
l=((i*i+i)/2);
for(int j=1;j<=2*i-1;j++)
{
if(i%2==0&&j%2==0)
{
printf("*");
}
else
{
printf("%ld",l);
l--;
k++;
}
}printf("\n");
}
}
}
how to print same in reverse way.
Deletelike 11*12*13*14*15
10*9*8*7
4*5*6
3*2
1
why are you cheking the conditions again.
Deleteif(i%2!=0&&j%2==0), no need of i%2!=0 becuase, you have entered the loop after cheking it only, same in else case also.
import java.io.*;
ReplyDeleteimport java.net.*;
public class Selenium_Projext {
public static void main(String[] args){
PrintPat(9);
}
public static void PrintPat(int a)
{ int last=0;
for(int i=1;i<=a;i++)
{ last=last+i;
int l=i;
for(int j=last;l>0;l--)
{
System.out.print(j--);
if(l>1)
System.out.print("*");
}
System.out.println();
}
}}
This comment has been removed by the author.
Deletepattern is not the same as expected
DeleteThanks for writing the same pattern in Java..!
ReplyDeleteThis comment has been removed by a blog administrator.
ReplyDeleteIn C (easy way)..
ReplyDelete#include
int main()
{
int i,j,n,count=0,k=0;
printf("Enter N");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
count=k;
for(j=1;j<=i;j++)
{
if(i%2==0)
{
printf("%d",count+i);
count=count-1;
if(j!=i)printf("*");
k++;
}
else
{
count=count+1;
printf("%d",count);
if(j!=i)printf("*");
k++;
}
}
printf("\n");
}
return 0;
}
import java.io.*;
ReplyDeleteimport java.net.*;
public class mindtree {
public static void main(String[] args)
{
PrintPat(5);
}
public static void PrintPat(int n)
{
int i,j,num=1,k=num,I=1;
for(i=1;i<=n;i++)
{
k=num-1;
for(j=1;j<=num;j++)
{
if(j%2==0)
System.out.print("*");
else
{
if(i%2==0)
{
System.out.print(k+I-num+i);
I++;
k=k-2;
}
else
System.out.print(I++);
}
}
num=num+2;
System.out.println();
}
This comment has been removed by the author.
ReplyDelete#include
ReplyDeleteint main(){
int i,j,k=1;
for(i=1;i<=5;i++){
for(j=1;j<=2*i-1;j++){
if(j%2==0){
printf("*");
}
else{
printf("%d",k);
k++;
}
}
printf("\n");
}
}
Your code is not for the pattern asked, its for different pattern
Deletecode in c language(easy to understand)
ReplyDelete#include
void main()
{
int i,j,k=1,n;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
if(i%2!=0)
{
for(j=1;j<=i;j++)
{
if(j<i)
printf("%d*",k++);
else
printf("%d",k++);
}
}
else
{
for(j=1;j<=i;j++)
{
if(j<i)
{
if(j==1)
k=k+i-1;
printf("%d*",k--);
}
else
printf("%d",k--);
if(j==i)
k=k+i+1;
}
}
printf("\n");
}
}
awesome man, this was one of the easiest way :)
Deletesuper! all test casses are passed.
DeleteNice post. Number Pattern Programs in Java .
ReplyDeletethanks
#include
ReplyDelete#include
using namespace std;
int main()
{
int i,j,k=1;
for(i=1;i<=5;i++)
{
if(i%2==0)
{
int temp=k-1+i;
for(j=1;j<=i*2-1;j++)
{
if(j%2==0)
{
printf("*");
}
else
{
printf("%d",temp);
temp--;
k++;
}
}
}
else
{
for(j=1;j<=2*i-1;j++)
{
if(j%2==0)
{
printf("*");
}
else
{
printf("%d",k);
k++;
}
}
}
printf("\n");
}
}
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeletePython 3.x code:
ReplyDeletecount,k = 0,0
n = int(input("enter n:"))
for i in range(1,n+1):
count = k
for j in range(1,i+1):
if i%2 == 0:
print(count+i,end="")
count = count - 1
if j!=i:
print("*",end="")
k += 1
else:
count = count +1
print(count,end="")
if j!=i:
print("*",end="")
k += 1
print("\r")
It doesn't take proper indentation when i'm posting it.
if anyone knows how to paste a code with proper indentation in this blog.please suggest me.
like in stackoverflow we do ctrl+k for code.
count,k = 0,0
ReplyDeleten = int(input())
for i in range(1,n+1):
count = k
for j in range(1,i+1):
if i%2 == 0:
print(count+i,end="")
count = count - 1
if j!=i:
print("*",end="")
k += 1
else:
count = count +1
print(count,end="")
if j!=i:
print("*",end="")
k += 1
print("\r")
count,k = 0,0
ReplyDeleten = int(input())
for i in range(1,n+1):
count = k
for j in range(1,i+1):
if i%2 == 0:
print(count+i,end="")
count = count - 1
if j!=i:
print("*",end="")
k += 1
else:
count = count +1
print(count,end="")
if j!=i:
print("*",end="")
k += 1
print("\r")
how to write java pattern program in below output
ReplyDelete1*2*3
7*8*9
4*5*6
1112
ReplyDelete3222
3334
how to write java pattern?
#include
Deleteint main()
{
int i,j,n;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
for(j=1;j<=n+1;j++)
{
if(i%2!=0)
{
if(j<=n)
{
printf("%d",i);
}
else
{
printf("%d",i+1);
}
}
else
{
if(j==1)
{
printf("%d",i+1);
}
else
{
printf("%d",i);
}
}
}
printf("\n");
}
return 0;
}
10*9*8*7
ReplyDelete6*5*4
3*2
1
Can u plz explain the logic?
ReplyDeleteC Program to print pattern
ReplyDeletenice post
ReplyDeleteC Program to print pattern