If you are looking to learn more about JUnit, go through these links that I hope will be useful for you.
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Hope this helps :)


I will try to explain the concept. First accept four integers input:

int num1,num2,num3,num4;

printf("Enter 4 integers seperated by space: ");
scanf("%d %d %d %d",num1,num2,num3,num4);

After this, do the function which accepts the two parameters.
For example, you are comparing two integers, num1 and num2:


In the function, have a if else statement that checks the two parameters. If num1 is more than num2, then put:

return num1;

This will return the biggest number. Assign the returned value to an int parameter. Put a printf statement if that is necessary. The printbiggest function will compare the four input using the if else statement and determine the biggest value of them. The function bigger() can be called in this function and assigned to an int variable. Then printf it.

on93 commented: thans +0


Complexity usually measured in terms of how the amount of a resource (time or space) required to run a program increases as the size of the program input increases.
Most common measure of complexity is the amount of time the program takes to run. If the algorithm of a program takes longer than the polynomial time(explained later), it is said to be intractable. If the algorithm takes lesser or within the polynomial time, it is said to be tractable.

O(1)         = constant
O(log a n)   = logarithmic
O(n)         = linear
O(n log a n) = “n log n”
O(n^2)       = quadratic
O(n^3)       = cubic
O(n^r)       = polynomial
O(a^n)       = exponential
O(n!)        = factorial

Anything that is larger than exponential is said to be exponential time algorithm or simply intractable. So, from constant to polynomial time algorithm can be classified as tractable while after polynomial can be classified as intractable. As the program goes down from constant to factorial, it has a reduced efficiency. I will provide an example of Big O notation.

(1) for(i = 0; i < n; i++)
(2)     for(k = 0; k < n; k++)
(3)        A[i,k] = 0;

Line (3) takes O(1) time

Line (2) takes O(n) time, since it executed according to the value of n. If the value of n is 5, the inner loop goes 5 times.

Line (1) takes O(n^2) time. If the the value of n is 2, the outer loop executes once and the inner loop ...