# function that finds factor def factors(n): print("Factors of",n,"are:") for i in range(1, n + 1): if n % i == 0: print(i) num = int(input('Enter the number: ')) print("The given number is "+str(num)) if(num >=0): factors(num) else: print("Invalid")
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- Write a function that takes in an integer n and computes n!. Do this without recursion. In [ ]: deffactorial_iter(n):"""Takes in an integer n>0 and returns the product of all integers from 1 to n."""# YOUR CODE HEREraiseNotImplementedError() In [ ]: In [ ]: assert factorial_iter(6) == 720 assert factorial_iter(7) == 5040 assert factorial_iter(10) == 3628800We rite a recursive function that calculates the products of natural numbers (1 through n) as follows: - asks the user to enter a positive number n - calculates and prints out the product of 1 through n (for example for n=5 the product is 120) Uploads: .c code and the outputHaving trouble with this c++ homework question. void a(int n) { if(n<1) { cout << n*2 << " "; return; } cout << n*2 << " "; a(n-1);}1. Trace the function when n is 4.
- Recursive Copy Function Given a string str and number n, write a program that recursively appends a copy of string str n times and returns the resulting string. Please complete the following function. n is the number of copies to add to str: function RecursiveCopy(str, n) {// your code here} The problem must satisfy the following: You must use recursion. You should end up with n+1 copies of the str total in the output because you are appending additional str copies. Use string literal formatting for the output.(Numerical) Write a program that tests the effectiveness of the rand() library function. Start by initializing 10 counters to 0, and then generate a large number of pseudorandom integers between 0 and 9. each time a 0 occurs, increment the variable you have designated as the zero counter; when a 1 occurs increment the counter variable that's keeping count of the 1s that occur; and so on. finally display the number of 0s, 1s, 2s, and so on that occurred and the percentage of the time they occurred.def cuberoot(n): """Computes the cube root of float n by fixpoint iteration""" cuberoot: This function should use fixed-point iteration to return the cube root of a given n (positive or negative floating point number). Your implementation should ideally return a precise approximation, but must be correct to at least eight significant digits and must use fixpoint iteration.
- iii. Add a function named lowercase() to this program that takes a list of uppercase letters as a parameter (e.g., x). This function must convert each uppercase letter in x to lowercase and return the list. Your function must use the ord() and chr() functions to solve this problem. You must NOT use the built-in lower() method. Hint: for any letter, the difference between its lowercase and uppercase ASCII values is 32. For example, ASCII values of 'a' and 'A' are 97 and 65, respectively; ASCII values of 'z' and 'Z' are 122 and 90, respectively.Given two integers x snf y, the following recurrsive definition determined the greatest common divisor of x and y, Write gcd(xy). Write a recursive function, gcd, that takes two integers as parameters and returns the greatest commmon divisorof numbers.Also write a program to test your function. Write a recursive function,reverseDigits, that takes an integer as a parameter snd returns the number with the digits reversed. Also write a program to test your application.3. Find Relatively Prime Write a function to return a number that is relatively prime with the given number. /* return an integer that is relatively prime with n, and greater than 1i.e., the gcd of the returned int and n is 1 Note: Although gcd(n,n-1)=1, we don't want to return n-1*/int RelativelyPrime (int n) 4. Find Inverse Then implement the following function, which returns the inverse modulo, and test it.Note that you need to check whether a and n are relative prime before calling thisfunction. Recall that a and n are relatively prime means that gcd(a,n)=1, i.e., theirgreatest common divisor is 1. Also recall that the extended Euclidean Algorithm canfind the integer s and t for a and n such that as+nt=gcd(a,n), where s is the inverseof a modulo n. /* n>1, a is nonnegative *//* a<=n *//* a and n are relative prime to each other *//* return s such that a*s mod n is 1 */int inverse (int a, int n){int s, t;int d = ExtednedEuclidAlgGCD (n, a, s, t);if (d==1){return (mod (t,…
- get_nth_word_from_string(s, n): This function takes a string s and a non-negative integer n as input. The string s may contain substrings of text separated by commas. The function should return the n’th comma-separated substring. We will start counting at 0, so if n=0, then the first substring should be returned (i.e., everything before the first comma, if there is such a comma); if n=1, then the second substring should be returned (i.e., everything between the first comma and second comma, if there is such a comma); and so on. If there is no n’th substring in s, then the empty string should be returned. Note: You may use the string count method in this function. Note: If a space follows a comma, do not include the space as part of the substring to be returned.get_nth_word_from_string(s, n): This function takes a string s and a non-negative integer n as input. The string s may contain substrings of text separated by commas. The function should return the n’th comma-separated substring. We will start counting at 0, so if n=0, then the first substring should be returned (i.e., everything before the first comma, if there is such a comma); if n=1, then the second substring should be returned (i.e., everything between the first comma and second comma, if there is such a comma); and so on. If there is no n’th substring in s, then the empty string should be returned. Note: You may use the string count method in this function. Note: If a space follows a comma, do not include the space as part of the substring to be returned. >>> get_nth_word_from_string("dreary, pondered, weak, weary", 0) 'dreary'2. Sum: a recursive function that computes the sum of integers 1, 2, 3, …., n for a given number n. So Sum(6) should return 1 + 2 + 3 + 4 + 5 + 6 , i.e. 21.sum(n) = n + sum(n-1)
