lower= int(input("lower range")) upper= int(input("upper range")) print("Odd numbers between 1 - 1000") def is_odd(n): if n%2!=0: return True else: return False def check_odds(i): for i in range(lower,upper+1): is_odd(i) print(i) I am trying to call is_odd function in check_odd to print odd numbers. please run this program to print correct output

lower= int(input("lower range")) upper= int(input("upper range")) print("Odd numbers between 1 - 1000") def is_odd(n): if n%2!=0: return True else: return False def check_odds(i): for i in range(lower,upper+1): is_odd(i) print(i) I am trying to call is_odd function in check_odd to print odd numbers. please run this program to print correct output

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter5: Control Structures Ii (repetition)

Section: Chapter Questions

Problem 14PE

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Similar questions

The sieve of Eratosthenes is a way of computing all the prime numbers below a certain number. (A prime number is a number that is only divisible by itself and 1). This algorithm is explained excellently in this video (Links to an external site.), or you can read about this ancient algorithm on Wikipedia (Links to an external site.). In python, implement this algorithm: Implement a function cross_out_multiples that takes as arguments a list of boolean values (true/false) called is_prime and a number n. The function sets the boolean values at all multiples of n (2*n, 3*n, 4*n ...) that are in the list to false. Implement a function sieve(n) which gives back a list of all primes below n. See template code attached.
Write a program called Super_Digit.java with a recursive function called superDigit. The function must return the calculated super digit as an integer. If 'x' has only 1 digit, then its super digit is 'x' Otherwise, the super digit of 'x' is equal to the super digit of the sum of the digits of 'x' Hint: use a combination of n%10 and n/10 The output should look like what is given below. Please give an editable code.
What is the output of the code segment below? Hint: Draw a 2x3 grid and then fill it in. int[][] plot = new int[2][3]; for (int s = 0; s < 2; s++) for (int t = 0; t < 3; t++) plot[s][t] = t - s + 3; for (int s = 0; s < 2; s++) { int sum = 0; for (int t = 0; t < 3; t++) sum += plot[s][t]; System.out.print(sum + " "); }
Write a Java function that counts the number of 'x's in the string. Any 'x' that is followed by another 'x' should count double (e.g. "axxbxc" -> 4) /** * Count the number of 'x's in the string. Any 'x' that is followed by * another 'x' should count double (e.g. "axxbxc" -> 4) * * @param x a string. * @return the count of x's. */ public static int countXWithDubs(String s){ return -1; }
As the algorithm performs division of numbers with high precision, it is very common to see really big numbers after the period (for example: 1200.2300001), which is not desirable because of several reasons: may cause confusion to some users when they see such big numbers; uses more memory to store a bigger number; it just does not make sense to display currency number in this format. For this reason, you are going to implement a utility function to format any number into the appropriate currency format, using 2 decimal places. For example: The number 1200.2300001 would be became: 1200.23
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In C++, an int is in the range 0 to 65535. But what if we need to add larger integers? Compute the sum 2345566777844567 + 9999988777765768009998. Write this function:string add(string a, string b) where a and b are strings expressing integers and the function add(a,b) returns a string signifying their sum. Strings a and b can have at most 100 characters. Write a main program that loops, asking the user for a and b. The program should terminate if the user enters “done” what asked for “number” a.
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Write a program with java that takes 10 numbers from user, then prints first largest and second largestnumbers (use loop and int type variable, do not use array).
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what sequence of numbers would be printed if the following function were executed with the value of N being 0? def xxx (N): print (N) if (N < 5): xxx (N + 2) print (N)
What will be the space complexity of the following function?public static double test (int n){double x;if (n == 0) return 1.0;else{x = 0.0;for (int i = 0; i < n; i++)x += test(i);return x;}}A. O(1)B. O(n)C. O(2^n)D. O(n!)