Skip to main content

Butterfly Pattern

Image
By Astrio Jather

 /* @ToDo 

Butterfly Pattern
    *                                   *
    *   *                           *   *
    *   *   *                   *   *   *
    *   *   *   *           *   *   *   *
    *   *   *   *   *   *   *   *   *   *
    *   *   *   *           *   *   *   *
    *   *   *                   *   *   *
    *   *                           *   *
    *                                   *
*/


#include <iostream>
using namespace std;

int main(){
     #ifndef ONLINE_JUDGE
        freopen("../asset/input.txt","r",stdin);
        freopen("../asset/output.txt","w",stdout);
    #endif
    // Code here!!

    int n; cin>>n;

    for(int i=1;i<=n;i++){
        for(int j=1;j<=2*n;j++){
            if(j<=i || j >(2*n)-i)
                cout<<"\t*";
            else cout<<"\t";
        }
        cout<<endl;
    }
    for(int i=1;i<=n;i++){
        for(int j=n-i;j>0;j--)
            cout<<"\t*";
        for(int j=2*i;j>0;j--)    
            cout<<"\t";
        for(int j=n-i;j>0;j--)
            cout<<"\t*";
        
        cout<<endl;
    }
    return 0;
}
Labels:

Comments

Post a Comment

Must Read:

Pyramid

By Astrio Jather
/*  To Print  1    2   2    3   3   3    4   4   4   4    5   5   5   5   5    */ #include   <iostream> using   namespace   std ; int   main (){       #ifndef  ONLINE_JUDGE          freopen ( "../asset/input.txt" , "r" , stdin );          freopen ( "../asset/output.txt" , "w" , stdout );     #endif     // Code here!!      int   n ;  cin >> n ;      for ( int   i = 1 ; i <= n ; i ++){          for ( int   j = 1 ; j <=  i ; j ++){     ...
Post a Comment
Read more

Zero One Knapsack | Recursion

By Astrio Jather
 1. You are given a number n, representing the count of items. 2. You are given n numbers, representing the values of n items. 3. You are given n numbers, representing the weights of n items. 3. You are given a number "cap", which is the capacity of a bag you've. 4. You are required to calculate and print the maximum value that can be created in the bag without       overflowing it's capacity. Note: Each item can be taken 0 or 1 number of times. You are not allowed to put the same item again and again. Input Format A number n v1 v2 .. n number of elements w1 w2 .. n number of elements A number cap Output Format A number representing the maximum value that can be created in the bag without overflowing it's capacity Constraints 1 <= n <= 20 0 <= v1, v2, .. n elements <= 50 0 < w1, w2, .. n elements <= 10 0 < cap <= 10 Sample Input 5 15 14 10 45 30 2 5 1 3 4 7 Sample Output 75 Solution: import java.io.*; import java.util.*; public class Main...
Post a Comment
Read more

Accenture Mock Quiz | Part 2

By Astrio Jather
  Question  10 Incorrect Marked out of 1.00 Flag question Question text Tom and his team uses the SVN(Configuration Management Tool) to do versioning of source code. Tom has pulled out the code from the SVN server to his local machine, updated his task partially and wanted to update the work back to the SVN server. What is the actual process Tom has performed with the SVN server. Select one: a.  fork/join   b.  check out/check in c.  commit/rollback d.  check in/check out Feedback The correct answer is: check out/check in Question  11 Incorrect Marked out of 1.00 Flag question Question text Software was developed for Global Marketing. Few changes that was earlier requested was already incorporated in the delivered software. Now the client does not want the changes and requested for the previous release. Which of the below option would facilitate developer to meet the client needs. Select one: a.  Software Control Management   b.  So...
Post a Comment
Read more

Mock 3 Slot3 - Handson - 21st July Java (30 Marks) 1. Farm Fest

By Astrio Jather
Anuj has a mango farm. He sells the mangoes and earns money from them. He fills the mangoes in a basket. To make the calculation much simpler, he fills the basket either with an odd number of mangoes or with an even number of mangoes. If the number of mangoes is a combination of both an even and an odd number of mangoes, then he needs to rearrange the basket. Note: Number of baskets cannot exceed 10 and cannot be zero or less, If so print “Invalid number of baskets”. A negative number of mangoes cannot exit, if so print "Invalid Input". Sample Input 1: Number of baskets 5 Number of mangoes in each basket 2 4 6 8 10 Sample Output 1: Even Sample Input 2: Number of baskets 4 Number of mangoes in each basket 13 29 45 5 Sample Output 2: Odd Sample Input 3: Number of baskets 3 Number of mangoes in each basket 12 13 14 Sample Output 3: Rearrange the mangoes Sample Input 4: Number of baskets 0 Sample Output 4: Invalid number of baskets Sample Input 5: Number of baskets 2 Number of ma...
Post a Comment
Read more

Pattern

By Astrio Jather
  /* To print Pattern:                     *                 *   *             *   *   *         *   *   *   *     *   *   *   *   * */ #include   <iostream> using   namespace   std ; int   main (){       #ifndef  ONLINE_JUDGE          freopen ( "../asset/input.txt" , "r" , stdin );          freopen ( "../asset/output.txt" , "w" , stdout );     #endif     //...
Post a Comment
Read more

Climb Stairs

By Astrio Jather
1. You are given a number n, representing the number of stairs in a staircase. 2. You are on the 0th step and are required to climb to the top. 3. In one move, you are allowed to climb 1, 2 or 3 stairs. 4. You are required to print the number of different paths via which you can climb to the top. Input Format A number n Output Format A number representing the number of ways to climb the stairs from 0 to top. Constraints 0 <= n <= 20 Sample Input 4 Sample Output 7 Solution: import java.io.*; import java.util.*; public class Main {     public static void main(String[] args) throws Exception {         // write your code here         Scanner sc = new Scanner(System.in);         int n = sc.nextInt();         System.out.println(WaysUp(n));     }     public static int WaysUp(int n){         if(n == 0)  return 0;         else if(n<0)...
Post a Comment
Read more

Programming using Java Hands On - Control Structures | Print Username

By Anonymous
  Print Username Jeffy, who is going to complete the higher education in this year needs to create a simple application which accept the name of a person and welcome them with a message along with their name. She wants to read the data using the class "Scanner". Implement this scenario using Java. Sample Input 1:   Enter the name: Johson Sample Output 1: Welcome Johnson. Sample Input 2:   Enter the name: Stain Polson Sample Output 2: Welcome Stain Polson. Result Description Summary of tests *Note: All the test cases might not have same weightage +------------------------------+ | 2 tests run/ 2 tests passed | +------------------------------+
Post a Comment
Read more

Climb Stairs With Variable Jumps

By Astrio Jather
 1. You are given a number n, representing the number of stairs in a staircase. 2. You are on the 0th step and are required to climb to the top. 3. You are given n numbers, where ith element's value represents - till how far from the step you       could jump to in a single move.        You can of course jump fewer number of steps in the move. 4. You are required to print the number of different paths via which you can climb to the top. Input Format A number n .. n more elements Output Format A number representing the number of ways to climb the stairs from 0 to top. Constraints 0 <= n <= 20 0 <= n1, n2, .. <= 20 Sample Input 10 3 3 0 2 1 2 4 2 0 0 Sample Output 5 Solution: import java.io.*; import java.util.*; public class Main {     public static void main(String[] args) throws Exception {         // write your code here         Scanner sc = new Scanner(System.in);   ...
Post a Comment
Read more

Logic Development | Object Oriented Programming Pre Quiz

By Astrio Jather
 
Post a Comment
Read more

Subscribe to Get's Answer by Email