Write a function that converts an input string to uppercase. The argument to the function is a string. The function then returns the string in uppercase. Use your function in a main program where you ask the user to enter a string and then you output the string in uppercase. Do not use the strupr or toupper functions. You have to use a loop. Hint: Characters in C are stored as integers using the American Standard Code for information interchange ( ASCII). The lower case characters are given the integers 97 to 122 where ‘a’=97, ‘b’=98,… The uppercase characters are represented by the integers 65 to 95 where ‘A’=65, ‘B’=66…If you have the statement putchar (97) in your program, then the character ‘a’ is output on the screen. Since characters are integers, then it is possible to do some basic mathematics with them e.g’a’+ 2=99 which is equal to c

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The Fibonacci series has the following form: 0, 1,1,2,3,5,8…. It can be solved recursively fib(n)=fib(n-1) + fib(n-2) i.e the nth fibonacci number is defined as shown b the above expression. Write a function using recursion that calculates the nth recursive number. For instance fib(3) = 1 i.e 1 + 0 and fib(4) = 2 i.e 1 + 1. Tip; The base case: fib(0)=0 and fib(1)=1. Use your function in a main program where you ask the user to a number n and then you output the nth Fibonacci number.

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Consider this prolog program rainy(vancouver) rainy(sept_iles). snowy(x) :- rainy(x), cold(x) Draw the search tree for the query snowy(C) The search begins at snowy(C) and finds the rule snowy(X), which then becomes the goal, with C unified to X. Prolog must then establish both rainy(X) and cold (X). It establishes rainy (X) by finding rainy(vancouver), thus (temporarily) unifying X with vancouver. But then it must also find cold(vancouver). Because this fails, prolog backtracks and so it unifies rainy with sep tiles. Going back up the tree, it finds cold(sep tiles) and returns snowy(sept iles).

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