Lab 1-Algorithm Function- designing and implementing algorithms using C program
60-141 – Introduction to Programming II
Objectives:
– Practice designing/implementing algorithms
– Practice use of functions
Code and document the following functions using NON-RECURSIVE ITERATION only.
Test the functions by calling them from a simple interactive main() function using a menu, with
different values used to select the choice of function. Overall, you should have one C program (call it Lab1.c) containing one main() function and 5 other functions, where the functions are called based on an interactive user menu:
Summation: Σ???=1=1+2+3+⋯+?
n > 1; reject with error message otherwise
[Note that this sum is equal to n(n+1)/2. DO NOT program the function – program the series.]
Factorial(0) = 1;
Factorial(n) = n * (n-1) * . . . * 2 * 1
Requirement: n >= 0; reject with error message otherwise
Fibonacci(0) = 0;
Fibonacci(1) = 1;
Fibonacci(n) = Fibonacci(n-1) + Fibonacci(n-2);
Requirement: n >= 0; reject with error message otherwise
gcd (x, y) = x, if y=0
gcd (x, y) = gcd (y, x MOD y), if y > 0
Requirement: x and y both > 0; reject with error message otherwise
Power(a,b) = ??
Requirement: a > 0, b > 0, b is an integer; reject with error message otherwise
Answer:
/* Title: Lab #1: Algorithm, Function Objective: Create a program with one main that solves 5 different problems using non-recursive method including Summation: (Σ1+2+3+...+n), factorial(n), fibonacci(n), gcd(x,y) and power(a,b) by calling a function using a menu. */ //Includes #include <stdio.h> //C-Preprocessor Directives //Function Prototypes int summation(int n); // The sum of n and all the integer numbers below it down to 0 int factorial(int n); // The product of n and all the integer numbers below it down to 1 int fibonacci(int n); // Fibonacci(n) returns the nth Fibonacci number. int gcd(int x,int y); // The greatest common denominator of the of the two integers numbers int power(int a,int b); //Integer a raised to the power of integer b //Main function int main(void){ int choice=0; //Main loop do{ // Display the menu choices to the user and prompt for input. printf("1- Use the SUMMATION with non-recursive iteration\n"); printf("2- Use the FACTORIAL with non-recursive iteration\n"); printf("3- Use the FIBONACCI with non-recursive iteration\n"); printf("4- Use the GCD with non-recursive iteration\n"); printf("5- Use the POWER with non-recursive iteration\n"); printf("0- Quit\n"); scanf("%d", &choice); switch(choice){ case 1:{ // Collect information and call summation(n) function. int n,result; printf("Please enter a value of n \n"); scanf("%d",&n); if(n<1){ printf("\nError, Please enter a number greater than or equal to 1\n\n"); continue; } else{ result=summation(n); printf("\nResult is %d \n\n", result); break;} } case 2:{ // Collect information and call factorial(n) function. int n,result; printf("Please enter a value of n \n"); scanf("%d",&n); if(n<0){ printf("\nError, Please enter a number greater than or equal to 0\n\n"); continue; } else{ result=factorial(n); printf("\nResult is %d \n\n", result); break; } } case 3:{ // Collect information and call fibonacci(n) function. int n,result; printf("Please enter a value of n \n"); scanf("%d",&n); if(n<0){ printf("\nError, Please enter a number greater than or equal to 0\n\n"); continue; } else{ result=fibonacci(n); printf("\nResult is %d \n\n", result); break; } } case 4:{ // Collect information and call gcd(x,y) function. int x,y,result; printf("Please enter a value for the first Number and then Enter a value for the Second Number \n"); scanf("%d %d",&x,&y); if(x<0 || y<0){ printf("\nError, Please enter a number greater than or equal to 0\n\n"); continue; } else{ result=gcd(x,y); printf("\nResult is %d \n\n", result); break; } } case 5:{ // Collect information and call power(x,y) function. int x,y,result; printf("Please enter a value for the first Number and then Enter a value for the Second Number \n"); scanf("%d %d",&x,&y); if(x<0 || y<0){ printf("\nError, Please enter a number greater than or equal to 0\n\n"); continue; } else{ result=power(x,y); printf("\nResult is %d \n\n", result); break; } } case 0:{ // Display a message for user after pressing 0 to Quit the program. printf("\nThank you for using this program, Please come back again \n\n\n"); break; } default:{ // Display a message for user after pressing a number other than 0,1,2,3,4 and 5. printf("\nIncorrect input, please choose a number between 1 and 5,or press 0 to Quit \n\n\n"); break; } } }while (choice !=0); // End the while loop if the choice was 0 return 0; //exit main }// end of main /* Objective: Compute the sum of n + all the numbers below it to 1 input: A positive integer that is 1 and higher output: The sum of n plus all integer numbers down to 1, example: ∑ n +(n-1) +...+ 4+ 3+ 2+ 1 */ int summation(int n){ int total=0,i; for (i = 1;i <=n;i++){ total += i; } return total; } /* Objective: Compute the factorial for integer n input: A positive integer that is 1 and higher (Non-negative integer) output: Return the factorial for integer n example: n*(n-1)*...* 4*3*2*1 */ int factorial(int n){ int i; int total = 1; for (i = 1; i <= n; i++){ total *= i; } return total; } /* Objective: Compute the nth Fibonacci number. input: A positive integer that is 1 and higher output: Compute the nth Fibonacci number which is determined by adding the last two Fibonacci numbers together, example: (to find the third Fibonacci number we add the two previous numbers in the sequence 0,1,1 which is 1+1=2) */ int fibonacci(int n) { int i, k = 0, j = 1, fib = 1; for (i = 1; i < n; i++){ fib = k + j; k = j; j = fib; } return fib; } /* Objective: Compute the greatest common denominator of the of the two integers numbers x and y input: x , y are integers and y is greater than 0 output: Return the greatest common denominator of the of the two integer numbers x,y such that(gcd(x,y)=gcd(y,x MOD y), if y>0 and gcd(x,y)=x, if y=0) */ int gcd(int x, int y){ int temp; while (y > 0){ temp = x; x = y; y = temp%y; } return x; } /* Objective: Compute a raised to the power of b input: Two positive integer a and b output: return a raised to the power of b. (example for a = 3, b = 3 then power(3,3)= a*a*a = 3*3*3=27). */ int power(int a, int b){ int power = 1, i; for (i = 0; i < b; i++){ power *= a; } return power; }
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