can you make a flow chart of this: #include int main(){ int num1, num2; for(num1 = 1; num1 <= 10; num1++){ printf("%d -- %d\n", num1, num2 = num1 * 2); } }
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can you make a flow chart of this:
#include <stdio.h>
int main(){
int num1, num2;
for(num1 = 1; num1 <= 10; num1++){
printf("%d -- %d\n", num1, num2 = num1 * 2);
}
}
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Solved in 2 steps with 1 images
- int const MULTIPLIER = 5; is a valid way to declare a constant integer variable. Group of answer choices True False ------- Given the following code segment, what is output to the screen? int num1 = 6; int num2 = 4 * num1++; cout << "num1=" << num1 << " num2=" << num2; Group of answer choices num1=7 num2=24 num1=6 num2=24 num1=6 num2=28 Nothing, because the code does not compile.In C++ write a program that generates three random numbers and then find the min number among the generated values, using these three functions: void getrandnum(int &n1, int &n2, int &n3), int findMin(int n1, int n2, int n3), and void printResult(int n1, int n2, int n3, int min). Make the main function drive all these functions.(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.
- In this assignment you will write your own square root function and a main program to test it. Thesquare root function will return a tuple the includes the square root of the number as well as thenumber of iterations it took to arrive at that solution. The pseudocode for the function and mainprogram is given as follows:import mathEPSILON = 1.0e-14def square_root(number):approximation = 1previous = 0iterations = 0while ((absolute value of (approximation * approximation minus number)is greater than EPSILON) and previous != approximation):previous = approximationapproximation = 0.5 * (approximation + number / approximation)iterations += 1return (approximation, iterations)print('This program calculates the square root of an inputted number')print('and the number of iterations that the algorithm took to reach')print(f'a degree of accuracy of {EPSILON}.')answer = “yes”while answer[0] equals “Y” or answer[0] equals “y”:number = float(input('Enter a nonnegative number: '))while number is less…Write a function named printN that takes an int n, a string s, and returns void. This function must print the string s, n times. void printN(int n, string s);cin >> n >> s; cout << n << s << endl; } Please help me with this problem using c++.In C, Given an int variable n that has already been declared, write some code that repeatedly reads a value into n until at last a number between 1 and 10 (inclusive) has been entered.
- Write a function that prints a diagonal line using asterisk characters. Your function should accept one parameter that determines the number of rows defining the length of the diagonal line. This value will always be a positive integer greater than or equal to 1 (e.g. 1, 2, 3, …). use this Java: public static void printPattern( int num );Does this function work? For example: $16.65 has 66 quarters, 1 dime, 1 nickel, and 0 pennies. If it doesn't work can you explain what the problem is. def coins(total, denomination): num = int(total / denomination) print(num, " ", end="") if denomination == 0.25: print("quarters") elif denomination == 0.10: print("dimes") elif denomination == 0.05: print("nickels") else: print("pennies") return total - (num * denomination)#----main----money = float(input("How much money do you have? > "))print()print("Your money in coins would be:")money = coins(money, 0.25)money = coins(money, 0.10)money = coins(money, 0.05)money = coins(money, 0.01)When we say arithmetic expression, we mean a kind of expression you can type in a very basic desk calculator. That is, neither variables nor brackets are allowed. It contains only numbers and four arithmetic operators, +, -, *, and /. The following are some examples of arithmetic expression: Any number string with or without signs - e.g. 3, -1, +10, 3.14, -0.70 and 099. Number strings mixed with arithmetic operators - e.g. 3+5, -1+2*3, 7/10-0.7, and -1.4-+8.2. An example of FA diagram for recognising a non-negative integer is given below. You may use it as a start point. You need to add a few more states and transitions to handle numbers with decimal point and signs (e.g. -5, +2, 0.21, -32.6, +99.05, but not the form 1.0E-3). Most importantly, you also need to add a few more things to deal with the arithmetic operators +, -, *, /.
- Write the following function to draw a regular polygon:def drawPolygon(x = 0, y = 0, radius = 50, numberOfSides = 3):The polygon is centered at (x, y) with a specified radius for the bounding circle for the polygon and the number of sides. Write a test program that displays a triangle, square, pentagon, hexagon, heptagon, and octagon, as shown in Figure 6.12a.Problem Description Did you know that in lotteries, a 3-digit number with the same numbers in all digits like 777 will hit the jackpot in a casino? In the same manner, let's make a program that will test if a certain 3-digit number hits a jackpot or not by identifying if all the digits of a given number is the same as the second inputted number. If it is, print "Jackpot!"; else, print "Nah". Input A line containing two integers separated by a space. 777·7 Output A line containing a string. Jackpot! -USED JAVA PROGRAMMING LANGUAGEI am trying to understand what it means when a variable or a function is static, but still not sure as to what it is. I understand this theoratically, but not sure as to how to when I should use this and how to apply this even in OOP when using the :: symbol? For example class Box{ static string set_name(string n); }; main.cpp #include "Box.h" #include <iostream> int main(){ cout << Box::set_name("my_name") << endl; }