You are given an integer money denoting the amount of money (in dollars) that you have and another integer children denoting the number of children that you must distribute the money to.
You have to distribute the money according to the following rules:
1 dollar.4 dollars.Return the maximum number of children who may receive exactly 8 dollars if you distribute the money according to the aforementioned rules. If there is no way to distribute the money, return -1.
Example 1:
Input: money = 20, children = 3 Output: 1 Explanation: The maximum number of children with 8 dollars will be 1. One of the ways to distribute the money is: - 8 dollars to the first child. - 9 dollars to the second child. - 3 dollars to the third child. It can be proven that no distribution exists such that number of children getting 8 dollars is greater than 1.
Example 2:
Input: money = 16, children = 2 Output: 2 Explanation: Each child can be given 8 dollars.
Constraints:
1 <= money <= 2002 <= children <= 30program main
implicit none
integer :: money, children, result
money = 20
children = 3
result = solve(money, children)
write (*, '(A, I0)') 'Result: ', result
money = 16
children = 2
result = solve(money, children)
write (*, '(A, I0)') 'Result: ', result
contains
function solve(money, children) result(result)
implicit none
integer, value :: money, children
integer :: i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
integer :: result
! This solution uses a brute force approach to find the maximum number of children
! who may receive exactly 8 dollars if we distribute the money according to the
! aforementioned rules.
!
! We start by considering all possible combinations of children who may receive 8 dollars.
! For each combination, we check whether it satisfies the rules. If it does, we increment
! the result variable.
!
! The time complexity of this solution is O(n^2), where n is the number of children.
! The space complexity is O(1).
result = -1
do i = 1, children
do j = 1, children
if (i + j == 8) then
if (i /= 1 .and. j /= 1) then
result = max(result, i + j)
end if
end if
end do
end do
do i = 1, children
do j = 1, children
do k = 1, children
if (i + j + k == 8) then
if (i /= 1 .and. j /= 1 .and. k /= 1) then
result = max(result, i + j + k)
end if
end if
end do
end do
end do
do i = 1, children
do j = 1, children
do k = 1, children
do l = 1, children
if (i + j + k + l == 8) then
if (i /= 1 .and. j /= 1 .and. k /= 1 .and. l /= 1) then
result = max(result, i + j + k + l)
end if
end if
end do
end do
end do
end do
do i = 1, children
do j = 1, children
do k = 1, children
do l = 1, children
do m = 1, children
if (i + j + k + l + m == 8) then
if (i /= 1 .and. j /= 1 .and. k /= 1 .and. l /= 1 .and. m /= 1) then
result = max(result, i + j + k + l + m)
end if
end if
end do
end do
end do
end do
end do
do i = 1, children
do j = 1, children
do k = 1, children
do l = 1, children
do m = 1, children
do n = 1, children
if (i + j + k + l + m + n == 8) then
if (i /= 1 .and. j /= 1 .and. k /= 1 .and. l /= 1 .and. m /= 1 .and. n /= 1) then
result = max(result, i + j + k + l + m + n)
end if
end if
end do
end do
end do
end do
end do
end do
do i = 1, children
do j = 1, children
do k = 1temp.f95:109:24:
109 | do k = 1
| 1
Error: Syntax error in iterator at (1)
f951: Error: Unexpected end of file in ‘temp.f95’
!-------------------------------------------------------------------------------
! This is a Fortran implementation of the problem described in the problem statement.
! The program takes two command line arguments: money and children.
! The program solves the problem and prints the maximum number of children who
! can receive exactly 8 dollars.
!-------------------------------------------------------------------------------
program max_children_8
implicit none
integer :: money, children, max_children
max_children = solve_problem(money, children)
write (*,*) max_children
contains
!---------------------------------------------------------------------------
! This function solves the problem by using dynamic programming.
! The function takes two arguments: money and children.
! The function returns the maximum number of children who can receive exactly
! 8 dollars.
!---------------------------------------------------------------------------
integer function solve_problem(money, children) result(max_children)
implicit none
integer, intent(in) :: money, children
integer, dimension(money, children) :: dp
integer :: i, j
dp(1,:) = 1
do i = 2, money
do j = 1, children
if (j == 1) then
dp(i,j) = 1
else
dp(i,j) = max(dp(i-1,j-1), dp(i-4,j))
end if
end do
end do
max_children = dp(money, children)
end function solve_problem
end program max_children_8
0
def maximum_children_with_eight_dollars(money, children):
if money < children * 1 or money > children * 8:
return -1
max_eights = (money - children) // 7
return children if max_eights >= children else max_eights
In this problem, we calculate the maximum number of children who get exactly 8 dollars by making sure that we follow the given rules. We first check if it's possible to distribute money with these constraints. If the amount of money is less than the product of the number of children and 1 or greater than the product of the number of children and 8, return -1 as it's not possible.
Then, find the maximum number of eights by subtracting the number of children from the amount of money and dividing the result by 7. We divide the result by 7, as we're trying to maximize the children getting exactly 8 dollars, and the other amounts could range from 1 to 3 or 5 to 7.
Finally, if the maximum number of eights is greater than or equal to the number of children, return the number of children. Otherwise, return the maximum number of eights. This calculation ensures that we're getting the maximum number of children who get exactly 8 dollars.
int maximumChildrenWithEightDollars(int money, int children) {
if (money < children * 1 || money > children * 8) return -1;
int max_eights = (money - children) / 7;
return max_eights >= children ? children : max_eights;
}
In this problem, we calculate the maximum number of children who get exactly 8 dollars by making sure that we follow the given rules. We first check if it's possible to distribute money with these constraints. If the amount of money is less than the product of the number of children and 1 or greater than the product of the number of children and 8, return -1 as it's not possible.
Then, find the maximum number of eights by subtracting the number of children from the amount of money and dividing the result by 7. We divide the result by 7, as we're trying to maximize the children getting exactly 8 dollars, and the other amounts could range from 1 to 3 or 5 to 7.
Finally, if the maximum number of eights is greater than or equal to the number of children, return the number of children. Otherwise, return the maximum number of eights. This calculation ensures that we're getting the maximum number of children who get exactly 8 dollars.