Download BIL106 Introduction to Scientific and Engineering Computing

BIL106 Introduction to Scientific and Engineering Computing
Sample Problems 2
Problem 1: Write a module containing a recursive function that calculates the sum of the first n
natural numbers, i.e. (S = 1 + 2 + 3 + …..+ n). Also write a main program to test the subroutine.
Solution 1:
module mproblem1
public :: total
contains
recursive function total(n) result(k)
integer, intent(in):: n
integer :: k
select case (n)
case (1)
k=1
case (2:)
k = n + total(n-1)
end select
end function total
end module mproblem1
! main program
program problem1
use mproblem1
integer :: i
print *, "Enter an integer number:"
read *, i
print *, "The sum of the first n natural
numbers is:"
print *, total(i)
end program problem1
Problem 2: Store 10 four digits numbers in an array. Purely by changing the output format, print
the numbers as
(a) a single column of numbers;
(b) three rows of four numbers;
(c) a single line of numbers;
(d) a single column of numbers in reverse order
(e) a single line of numbers in reverse order
Your anwers should be a single print statement. The numbers should neither overlap nor be
adjacent.
Solution 2:
program problem2
integer, dimension(12) :: arr
integer :: i
arr = (/(12345+500*i, i =1, 12)/)
print "(i5)", arr
print "(4(i5, tr1))", arr
print "(12(i5,tr1))", arr
print "(i5)", arr(12:1:-1)
print "(12(i5,tr1))", arr(12:1:-1)
end program problem2
Problem 3: Personal data of the employees of company X are stored in a file named
“phonenum.txt” as is given below:
Name
------John
Ashley
Mike
…………
…………
…………
Surname
-----------Doe
Long
Young
Phone number
----------------2343434
9871233
3178256
The file contains the data of 100 employees. Write a program that reads the data from this file then
puts them into alphabetical order with respect to the surnames and then stores the personal data in a
new file called “ordered.txt.”
Solution 3:
program problem3
character(len = 11), dimension(100) :: name, sname, tnum
character(len = 11) :: temp
integer:: i, j, n
open (unit = 10, file = "phonenum.txt", action = "readwrite", status = "old",
position="rewind")
n = 55
!dosyada sadece 55 veri bulunduğu için burada n = 55 alınmış.
do i = 1, n
read (unit = 10, fmt="(a10,a11,a7)") name(i), sname(i), tnum(i)
end do
do i = 1, n-1
do j = i+1, n
if (sname(i)>sname(j)) then
temp = sname(i)
sname(i) = sname(j)
sname(j) = temp
temp = name(i)
name(i) = name(j)
name(j) = temp
temp = tnum(i)
tnum(i) = tnum(j)
tnum(j) = temp
endif
end do
end do
open (unit = 11, file = "ordered.txt", action = "readwrite", status = "replace",
position="rewind")
do i = 1, n
write(unit = 11, fmt="(3a11)") name(i), sname(i), tnum(i)
end do
end program files
Problem 4: Write a subroutine that multiplies two n-by-n matrices A and B and assign the result to
an output array C. Also write a main program to test the subroutine.
Solution 4:
module mproblem4
public :: multpl
contains
subroutine multpl(a, b, c)
real, dimension(:, :), intent(in) :: a, b
real, dimension(:,:), intent(out) :: c
integer :: i, j, k, n
real :: summ
n = size(a,1)
do i = 1, n
do j = 1, n
summ = 0
do k = 1, n
summ = summ + a(i, k) * b(k, j)
end do
c(i, j) = summ
end do
end do
end subroutine multpl
end module mproblem4
!main program
program problem4
use mproblem4
real, dimension(3, 3) :: f, g, h
integer :: i, j
do i = 1, 3
do j = 1, 3
f(i,j) = (i - 1.0)* 3 + j * 1.0
g(i,j) = 9.0 + (i - 1.0) * 3.0 + j * 1.0
end do
end do
call multpl(f, g, h)
print *, "matrix F"
print *, "--------"
print "(3f5.1)", f
print *, " "
print *, "matrix G"
print *, "--------"
print "(3f5.1)", g
print *, " "
print *, "Multiplication of matrix F and G =
matrix H"
print *, "-------------------------------------------"
print "(3f6.1)", h
end program problem4
Problem 5: In mathematics, a perfect number is a positive integer that is the sum of its proper
positive divisors, that is, the sum of the positive divisors excluding the number itself. Equivalently,
a perfect number is a number that is half the sum of all of its positive divisors (including itself). The
first perfect number is 6, because 1, 2, and 3 are its proper positive divisors, and 1 + 2 + 3 = 6.
Equivalently, the number 6 is equal to half the sum of all its positive divisors:
(1 + 2 + 3 + 6) / 2 = 6.
(a) Write a function that determines whether an integer number is a perfect number and returns
the logical value .true. if it is.
(b) Write a main program that finds the perfect numbers lying in the interval [1, 10000].
Solution 5:
module mproblem6
public :: perfect
contains
function perfect(n) result(bool)
integer, intent(in) :: n
logical :: bool
integer :: summ, i
summ = 0
do i = 1, n/2
if (modulo(n,i)==0) then
summ = summ + i
end if
end do
program problem6
use mproblem6
integer :: i
do i = 1, 10000
if (perfect(i)) then
print *, i
end if
end do
end program problem6
if (summ ==n) then
bool = .true.
else
bool = .false.
end if
end function perfect
end module mproblem6