Description
Problem 1: Accumulator Machine
In this problem we try to mimic a simplified accumulator architecture. An accumulator machine is a type
of 1-operand machine where it uses an “accumulator” to store the results. For example, the add operation
would be something like “add R1” which means add the value of R1 to the accumulator, so if accumulator
was 0 now it will be equal to R1. If another command such as “add R2” is issued, then R2 will be added to
accumulator which has the value of R1 so that the new value of accumulator is R1 + R2.
You will be given the input command (operator) as a number between 0 and 9, followed by the number
(the operand) in each line of stdin (read with scanf). Assume accumulator starts at 0. EXIT, INC, DEC, PRINT
and RESET don’t have an operand. If the command is Print then you print the value of the accumulator
with a newline, if it is RESET, the value of accumulator is reset to 0. If it is EXIT you exit the program. INC
would increment the accumulator by 1 whereas DEC would decrement it by one.
In cases of division and modulus, the remainder must be always positive. You must make sure both
division and modulus results are consistent.
Sample:
Input:
1 50
2
5 3
8
7 100
8
9
1 100
3 10
8
0
Output:
153
53
90
0 1 2 3 4 5 6 7 8 9
EXIT ADD INC SUB DEC MULT DIV MOD PRINT RESET
Problem 2: Fibonacci Number
A number in the Fibonacci sequence is the one that is the sum of the two preceding numbers in the
sequence. For instance, if the first number is 5 and the second is 3 the third would be 8, fourth would be
11 and so on. For this problem you have to write a program where it takes in the first two numbers of the
sequence and another number N in a newline where you have to compute and print the Nth element of
the sequence.
Sample
Input1:
5 10
9
Output1:
275
Input2:
1 1
12
Output2:
144
Problem 3: Sacred number!
A number is sacred if it has an even number of zeroes and is divisible by 13 or it has odd number of zeroes
and is divisible by 7. Write a program to test whether a given number is sacred or not. Print 1 if it is sacred
and 0 otherwise.
If you just printf 1 or 0 for the output in hope of getting 50% (without implementing any logic) you will
not receive any points for the problem.
Samples:
Input1:
3003
Output1:
1
Input2:
5810
Output2:
1
Input3:
6320
Output3:
0
Input4:
6323
Output4:
0
Input5:
9893
Output5:
1