QUESTION 6 (5 pt): What is the output? #include void mystery (int x[I], int n) { int temp; for (int i temp =0; i < n / 2 - 1; x[i]; x[n - i - 1]; i++) { P x[i] x[n - i - 1] = temp; } } int main() int a[] = 1, 2, 3, 4, 9}; mystery (a, 4 ) ; for ( auto element : a) std::cout << element << endl; } 1 4 5 23
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- T/F 6. Consider the following recursive sum method:public int sum(int x){if (x == 0) return 0;else return sum(x - 1) + 1;}JAVA Question 2: For two integers m and n, their GCD (Greatest Common Divisor) can be computed by a recursive method. Write a recursive method gcd(m,n) to find their Greatest Common Divisor. Method body: If m is 0, the method returns n. If n is 0, the method returns m. If neither is 0, the method can recursively calculate the Greatest Common Divisor with two smaller parameters: One is n, the second one is m mod n (or m % n). The recursive method cannot have loops. Note: although there are other approaches to calculate Greatest Common Divisor, please follow the instructions in this question, otherwise you will not get the credit. main method: Prompt and read in two numbers to find the greatest common divisor. Call the gcd method with the two numbers as its argument. Print the result to the monitor. Example program run: Enter m: 12 Enter n: 28 GCD(12,28) = 4 And here is what I have so far, package CSCI1302;import java.util.*;public class RecursionDemo { public static void…Problem 4: Magic Square Test A magic square of order n is an arrangement of n × n numbers, usually distinct integers, in a square, such that the n numbers in all rows, all columns, and both diagonals sum to the same constant. See all rows’, columns’ and diagonal’s sum is same. Write a main method where first you need to input n. Then input n*n integers and form a 2D matrix. Print “YES” if the matrix is magic square and print “NO” Otherwise. Sample input: 3 2 7 6 9 5 1 4 3 8 Output: YES
- /** * Write a Java function to determine if the elements of an array can be split into two groups such that the sum of one group is a non-zero multiple of 10 and the sum of the other group is odd. Your solution must use recursive backtracking. <code> * Example: * Given: {1, 2, 6, 3, 1} * Result: true because {1,2,6,1}, {3} is one possibility. * * Example: * Given: {4, 3, 5, 2} * Result: false * </code> */ public static boolean splitOdd10(int[] num) { }Q#2 Write a recursive function zeroCount ( int a[ ], int s, int e) that receives an array of integers a [], a start index s, and an end index e. The function should return the number of zeros in that array between s and e. int zeroCount ( int a[ ], int s, int e); Trace your function given the following array and function call. Draw your steps. int a[ ] = {1, 0, 0, 5}; int zeros = zeroCount(a, 0, 3); language c++in c++ You are given an array A of non-negative integers of size m. Your task is to sort the array in increasing order and print out the original indices of the new sorted array. Example: A= {12, 2, 6, 10, 9, 24,23, 33} After sorting the new array becomes A= {2,6,9,10,12,23,24,33}. The required output should be sorted array and original indexes "1 2 4 3 0 6 7", index of 2 in original array was 1 , index for 6 was 2 and so on.
- Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for someone to come and tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The count of how many different ways a positive integer n can be represented as a sum of consecutive integers is also called its politeness, and can be alternatively computed by counting how many odd divisors that number has. However, note that the linked…Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for the rugged hero to tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The number of ways that a positive integer n can be represented as a sum of consecutive integers is called its politeness, and can also be computed by tallying up the number of odd divisors of that number. However, note that the linked Wikipedia de0inition…Problem 4.1: Implement a recursive method that takes as a parameter a non-negative integer and generates the following patterns of starts. If the non-negative integer is 5, then the pattern generated is ***** **** *** ** * * ** *** **** ***** Note: java programming please follow hint
- A matrix is a rectangle of numbers in rows and columns. A 1xN matrix has one row and N columns. An NxN matrix has N rows and N columns. Multiplying a 1xN matrix A and an NxN matrix B produces a 1xN matrix C. To determine the Nth element of C multiply each element of A by each element of the Nth column of B and sum the results. Helpful information can be found at matrix multiplication. Write a program in Java that reads a 1xN matrix A and an NxN matrix B from input and outputs the 1xN matrix product, C. The first integer input is N, followed by one row of N integers for matrix A and then N rows of N integers for matrix B. N can be of any size >= 2. For coding simplicity, follow each output integer by a space, even the last one. The output ends with a newline. Ex: If the input is: 2 2 3 1 2 3 4 A contains 2 and 3, the first row of B contains 1 and 2, and the second row of B contains 3 and 4. The first element of C is (2 * 1) + (3 * 3), and the second element of C is (2 * 2) + (3 * 4).…Need python help. For problems 1 and 2, add your code to the file Lab2.java. Add your tests in the main function of Lab2.java. Do not use static variables in class Lab2 to implement recursive methods. Problem 1: Implement a recursive method min that accepts an array and returns the minimum element in the array. The recursive step should divide the array into two halves and find the minimum in each half. The program should run in O(n) time and consume O(logn) memory. Demonstrate the output of min on the array int [] a = { 2, 3, 5, 7, 11, 13, 17, 19, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 23, 29, 31, 37, 41, 43 } Problem 2 You have been offered a job that pays as follows: On the first day, you are paid 1 cent, on the second day, 2 cents, on the third day, 4 cents and so on. In other words, your pay doubles every day. Write a recursive method computePay that for a given day number computes the pay in cents. Assume that you accumulate all the money that you are paid. Write a…1: |R|←|P| Reserve space for |P| = 13 values.2: x ← n3: for i ← 0 ...(|P| − 1) do4: c ← x div Pi Number of multiplicands Pi in x.5: Ri ← c6: x ← x − c · Pi7: end for8: return RA Java programmer could implement Algorithm by first modelling the primitivenumbers with the enumeration type RomanNumeral. Each enum constant (I, IV, ..., M) is declared with its decimal value, which can be accessed with the function getValue().