You are given a positive integer n. Each digit of n has a sign according to the following rules:
Return the sum of all digits with their corresponding sign.
Example 1:
Input: n = 521 Output: 4 Explanation: (+5) + (-2) + (+1) = 4.
Example 2:
Input: n = 111 Output: 1 Explanation: (+1) + (-1) + (+1) = 1.
Example 3:
Input: n = 886996 Output: 0 Explanation: (+8) + (-8) + (+6) + (-9) + (+9) + (-6) = 0.
Constraints:
1 <= n <= 109.spoilerbutton {display:block; border:dashed; padding: 0px 0px; margin:10px 0px; font-size:150%; font-weight: bold; color:#000000; background-color:cyan; outline:0; } .spoiler {overflow:hidden;} .spoiler > div {-webkit-transition: all 0s ease;-moz-transition: margin 0s ease;-o-transition: all 0s ease;transition: margin 0s ease;} .spoilerbutton[value="Show Message"] + .spoiler > div {margin-top:-500%;} .spoilerbutton[value="Hide Message"] + .spoiler {padding:5px;}
program main
implicit none
integer, parameter :: n = 521
integer :: i, j, k, l, m, n_digits
character(len=1) :: digit
logical :: negative
! Count the number of digits in n
n_digits = 0
do while (n /= 0)
n_digits = n_digits + 1
n = n / 10
end do
! Print the sum of all digits with their corresponding sign
write (*, '(A)') 'Sum of all digits with their corresponding sign:'
do i = 1, n_digits
! Get the next digit from n
digit = char(ichar('0') + mod(n, 10))
n = n / 10
! Determine the sign of the current digit
if (i == 1) then
! Most significant digit is always positive
negative = .false.
else
! Other digits have opposite signs as their adjacent digits
negative = (mod(i, 2) == 0)
end if
! Print the digit with its sign
if (negative) then
write (*, '(A)', advance='no') '-'
else
write (*, '(A)', advance='no') '+'
end if
write (*, '(A)', advance='no') digit
end do
write (*, *)
end program maintemp.f95:13:8:
13 | n = n / 10
| 1
Error: Named constant ‘n’ in variable definition context (assignment) at (1)
temp.f95:21:8:
21 | n = n / 10
| 1
Error: Named constant ‘n’ in variable definition context (assignment) at (1)
PROGRAM SUM_DIGITS
IMPLICIT NONE
INTEGER :: n, sum, digit, i
CHARACTER(LEN=1) :: sign
! Read input
READ(*,*) n
! Initialize sum and sign
sum = 0
sign = '+'
! Iterate over digits of n
DO i = 1, LEN_TRIM(n)
! Get digit and sign
digit = INT(n(i:i))
IF (i == 1) THEN
sign = '+'
ELSE
sign = '-'
END IF
! Add digit to sum
sum = sum + digit * sign
END DO
! Output result
WRITE(*,*) sum
END PROGRAM SUM_DIGITS
temp.f95:17:23:
17 | digit = INT(n(i:i))
| 1
Error: Syntax error in argument list at (1)
temp.f95:25:28:
25 | sum = sum + digit * sign
| 1
Error: Operands of binary numeric operator ‘*’ at (1) are INTEGER(4)/CHARACTER(1)
temp.f95:15:25:
15 | DO i = 1, LEN_TRIM(n)
| 1
Error: ‘string’ argument of ‘len_trim’ intrinsic at (1) must be CHARACTER
def sum_of_digits(n: int) -> int:
total = 0
positive = True
while n > 0:
digit = n % 10
n //= 10
if positive:
total += digit
else:
total -= digit
positive = not positive
return total
The algorithm is simple and works as follows:
1. Initialize a variable, sum, to store the sum of the digits with their signs, and a boolean variable, positive, to determine whether the current digit should have a positive or negative sign.
2. Iterate through the digits of the input number n by repeatedly dividing it by 10 and checking if it's greater than 0.
- Note that we process the digits from the least significant digit to the most significant digit.
3. At each iteration, get the least significant digit (digit) of n by taking the remainder of the division of n by 10.
4. If positive is true, add the digit to the sum; otherwise, subtract it from the sum.
5. Toggle the value of positive for the next iteration.
6. Return the final value of sum.
The algorithm has a time complexity of O(log(n)) because it processes each digit of the input number n.
int sumOfDigits(int n) {
int sum = 0;
bool positive = true;
while (n > 0) {
int digit = n % 10;
n /= 10;
if (positive) {
sum += digit;
} else {
sum -= digit;
}
positive = !positive;
}
return sum;
}
The algorithm is simple and works as follows:
1. Initialize a variable, sum, to store the sum of the digits with their signs, and a boolean variable, positive, to determine whether the current digit should have a positive or negative sign.
2. Iterate through the digits of the input number n by repeatedly dividing it by 10 and checking if it's greater than 0.
- Note that we process the digits from the least significant digit to the most significant digit.
3. At each iteration, get the least significant digit (digit) of n by taking the remainder of the division of n by 10.
4. If positive is true, add the digit to the sum; otherwise, subtract it from the sum.
5. Toggle the value of positive for the next iteration.
6. Return the final value of sum.
The algorithm has a time complexity of O(log(n)) because it processes each digit of the input number n.