print_shampoo_instructions() Write a function with parameter num_cycles. If num_cycles is less than 1, print "Too few.". If more than 4, print "Too many.". Else, print "N: Lather and rinse." num_cycles times, where N is the cycle number, followed by "Done." Sample output with input: 2 1 Lather and rinse. 2 Lather and rinse. Done. Hint: Define and use a loop variable. 391908.2626244.qx3zqy7
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- Q6 In Python implement a recursive function named test() that accepts an integer parameter. If the parameter value is even then the test function should divide the parameter value by 2 and return this value. If the parameter value is odd, then the function should return 3 times the parameter value + 1. Then request a user to enter an integer number and recursively call converge() on that number until the function returns the value 1.Implement the sieve of Eratosthenes: a function for computing prime numbers, known to the ancient Greeks. Choose an integer n. This function will compute all prime numbers up to, and including, n. First insert all numbers from 1 to n into a set. Then erase all multiples of 2 (except 2); that is, 4, 6, 8, 10, 12, ... . Erase all multiples of 3, that is, 6, 9, 12, 15, ... . Go up to ?⎯⎯√n. The remaining numbers are all primes. Make sure that you use functions in your solution, including a main function. Define def sieveOfEratosthenes(n) that receives the maximum value and returns a list of all prime numbers less than or equal to this value in main() print the list of prime numbers.in c++ 1. Write a function that takes a 1 Dimensional array and an integer n andreturns the number of times ‘n’ appears in the array. If ‘n’ does not appearin the array, return -1.2. Write a function that takes a 2 Dimensional array and returns the positionof the first row with an odd sum. Assume that the column size is fixed at 4.If no sum is odd, return -13. Write a class, “pie”, that has a number of slices (int slices) as a privateproperty. Construct the pie with a number of slices and remove a slice witha function. Tell the user how many slices are in the pie.
- 12. Consider the recursive function int gcd( int a, int b) int gcd( int a, int b){ if (b > a) return gcd(b,a); if ( b == 0 ) return a; return gcd( b, a% b); } How many invocation (calls) of the gcd() function will be made by the call gcd(72, 30)?Implement the design of the Student_Tutor class that inherit from Student class so that the following code generates the output below:Hint: Each room has a capacity of 2 Student tutors.class Student: def __init__(self,name,student_id): self.name = name self.student_id = student_id self.semester = "Spring 2022" def details(self): print("Name:",self.name) print("Student_ID:",self.student_id) print("Current Semester:",self.semester)# Write your code here:STs = []S1 = Student("Reem", 18101235)S1.details()print("1.======================================")ST1 = Student_Tutor("Moynul", 16107892, 35, "CSE110", "CSE111")ST2 = Student_Tutor("Priyo", 16107792, 25,"CSE320")ST3 = Student_Tutor("Ronok", 18207892, 36, "CSE110", "CSE111", "CSE221", "CSE220")ST4 = Student_Tutor("Zeba", 19307892, 48, "CSE110")STs.extend([ST1, ST2, ST3, ST4])print("2.======================================")for i in range(len(STs)): STs[i].details()…Write a recursive function int fib (int n) to compute the Fibonacci numbers where n is a positive integer.Write an application (main() function) using command line parameters that calls fib(n) and outputs n,fin(n), fib(n)*1.0/fib(n-1) for n = 3,4,5,6,7,8,9,10,11,12,13,14,15In a format fprintf (stdout,“n = %d\tfib(n) = %d\tfib(n)/fib(n-1) = %.4f\n”, n, fib(n), fib(n)*1.0/fib(n-1))The ratio of fib(n) to fib(n-1) for large values of n (larger than say 10) is called the golden ratio.Plot a curve between n and golden ratio.Submit the assignment by dropping the MS word file containing the program code, results (take a screenshot of the Cygwin/ubantu) and paste to the word file in C
- Implement the function below. void swap(int pos1, int pos2){} Initial code to be completed: class ArrayList : public List { int* array; int index; int capacity; void dyn_all_add(){ int cap = ceil(capacity * 1.5); array = (int*)realloc(array,cap * sizeof(int)); capacity = cap; } void dyn_all_rem(){ int cap = capacity - (capacity/3); array = (int*)realloc(array,cap * sizeof(int)); capacity = cap; } public: // CONSTRUCTOR ArrayList() { capacity = 4; array = (int*)malloc(capacity); index = 0; } int add(int num) { if (index == capacity){ dyn_all_add(); } *(array + index) = num; index++; return index; } int get(int pos){ if (pos-1 < index){ return *(array + pos-1); } return -1; } int size(){ return index; }…TODO 1 Obtain all the indexes with labels equal 2 with np.where and the label array y. Keep the results in two_class_idx also index np.where() at 0 # TODO 1.1 two_class_idx = print(f"two_class_idx output: \n {two_class_idx}") try: print(f"two_class_idx shape: {two_class_idx.shape}") except Exception: pass todo_check([ (isinstance(two_class_idx, np.ndarray),f'two_class_idx is not an NumPy array! two_class_idx is currently a {type(two_class_idx)}'), (np.all(two_class_idx == np.array([3,9,11,12,16])),'two_class_idx does not contain the correct location values') ])2. Implement a function called findMinGap that will return the smallest gap betweenadjacent entries of an integer array. A gap between two numbers is the absolutevalue of their difference. For example, if an array contains the elements {10, 14,-5, -3, 0, 5, 7}, the minimum gap is 2 (between -5 and -3).Title line: int findMinGap(int array[], int length)
- 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, write a function that gets 3 pointers int* a, int* b, int* c, and rotates the values in their addresses to the left. That is, a gets the value of b, b gets the value of c, and c gets the value of a. void rotate3 (int* a, int* b, int* c); For example, if we have int x=1, y=2, z=3, then after calling rotate3 (&x, &y, &z) we should have x==2, y==3, and z==1. if we have int x=7, y=1, z=6, then after calling rotate3 (&x, &y, &z) we should have x==1, y==6, and z==7.Implement the FindText() function, which has two strings as parameters. The first parameter is the text to be found in the user provided sample text, and the second parameter is the user provided sample text. The function returns the number of instances a word or phrase is found in the string. In the PrintMenu() function, prompt the user for a word or phrase to be found and then call FindText() in the PrintMenu() function. Before the prompt, call cin.ignore() to allow the user to input a new string.Ex: Enter a word or phrase to be found: more "more" instances: 5