Given two non-negative integers, num1 and num2 represented as string, return the sum of num1 and num2 as a string.
You must solve the problem without using any built-in library for handling large integers (such as BigInteger). You must also not convert the inputs to integers directly.
Example 1:
Input: num1 = "11 ", num2 = "123 " Output: "134 "
Example 2:
Input: num1 = "456 ", num2 = "77 " Output: "533 "
Example 3:
Input: num1 = "0 ", num2 = "0 " Output: "0 "
Constraints:
1 <= num1.length, num2.length <= 104num1 and num2 consist of only digits.num1 and num2 don't have any leading zeros except for the zero itself.program main
implicit none
character(len=100) :: num1, num2, result
! Example 1
num1 = "11 "
num2 = "123 "
call add_strings(num1, num2, result)
print '(A)', trim(result)
! Example 2
num1 = "456 "
num2 = "77 "
call add_strings(num1, num2, result)
print '(A)', trim(result)
! Example 3
num1 = "0 "
num2 = "0 "
call add_strings(num1, num2, result)
print '(A)', trim(result)
contains
subroutine add_strings(num1, num2, result)
implicit none
character(len=*), intent(in) :: num1, num2
character(len=*), intent(out) :: result
integer :: i, j, carry
result = ""
carry = 0
do while (i < len(num1) .or. j < len(num2))
if (i < len(num1)) then
carry = carry + ichar(num1(i+1:i+1)) - ichar('0')
end if
if (j < len(num2)) then
carry = carry + ichar(num2(j+1:j+1)) - ichar('0')
end if
result = char(carry%10 + ichar('0')) // result
carry = carry / 10
i = i + 1
j = j + 1
end do
if (carry > 0) then
result = char(carry%10 + ichar('0')) // result
end if
end subroutine add_strings
end program maintemp.f95:41:33:
41 | result = char(carry%10 + ichar('0')) // result
| 1
Error: Unexpected ‘%’ for nonderived-type variable ‘carry’ at (1)
temp.f95:47:33:
47 | result = char(carry%10 + ichar('0')) // result
| 1
Error: Unexpected ‘%’ for nonderived-type variable ‘carry’ at (1)
PROGRAM SUM_STRINGS
IMPLICIT NONE
CHARACTER(len=1000) :: num1, num2, result
INTEGER :: i, j, carry
! Read the input numbers
READ(*,*) num1, num2
! Initialize the result string
result = "0"
! Initialize the carry
carry = 0
! Sum the digits of the two numbers
DO i = LEN(num1), 1, -1
DO j = LEN(num2), 1, -1
IF (i .LE. LEN(num1) .AND. j .LE. LEN(num2)) THEN
! Add the current digits and the carry
result(i:i) = CHAR(MOD(INT(result(i:i)) + INT(num1(j:j)) + carry, 10) + 48)
carry = INT(result(i:i)) + INT(num1(j:j)) + carry
ELSE IF (i .LE. LEN(num1)) THEN
! Add the current digit and the carry
result(i:i) = CHAR(MOD(INT(result(i:i)) + carry, 10) + 48)
carry = INT(result(i:i)) + carry
ELSE IF (j .LE. LEN(num2)) THEN
! Add the current digit and the carry
result(i:i) = CHAR(MOD(INT(num2(j:j)) + carry, 10) + 48)
carry = INT(num2(j:j)) + carry
END IF
END DO
END DO
! Print the result
WRITE(*,*) result
END PROGRAM SUM_STRINGS
This program uses the `MOD` function to perform the addition of the digits of the two numbers, and the `INT` function to convert the characters to integers. The `LEN` function is used to get the length of the strings. The `CHAR` function is used to convert the integers to characters. The `READ` statement is used to read the input numbers from the console, and the `WRITE` statement is used to print the result to the console.
Note that this program assumes that the input numbers are non-negative integers represented as strings. If the input numbers can be negative integers, then the program will need to be modified to handle the negative sign.
Also note that this program does not use any built-in library for handling large integers, and it does not convert the inputs to integers directly. It uses the `MOD` function to perform the addition of the digits of the two numbers, and the `INT` function to convert the characters to integers.
This program should run with all provided examples and output to stdout.
temp.f95:40:132:
40 | This program uses the `MOD` function to perform the addition of the digits of the two numbers, and the `INT` function to convert the characters to integers. The `LEN` function is used to get the length of the strings. The `CHAR` function is used to convert the integers to characters. The `READ` statement is used to read the input numbers from the console, and the `WRITE` statement is used to print the result to the console.
| 1
Error: Line truncated at (1) [-Werror=line-truncation]
temp.f95:42:132:
42 | Note that this program assumes that the input numbers are non-negative integers represented as strings. If the input numbers can be negative integers, then the program will need to be modified to handle the negative sign.
| 1
Error: Line truncated at (1) [-Werror=line-truncation]
temp.f95:44:132:
44 | Also note that this program does not use any built-in library for handling large integers, and it does not convert the inputs to integers directly. It uses the `MOD` function to perform the addition of the digits of the two numbers, and the `INT` function to convert the characters to integers.
| 1
Error: Line truncated at (1) [-Werror=line-truncation]
temp.f95:21:61:
21 | result(i:i) = CHAR(MOD(INT(result(i:i)) + INT(num1(j:j)) + carry, 10) + 48)
| 1
Error: ‘a’ argument of ‘int’ intrinsic at (1) must have a numeric type
temp.f95:22:46:
22 | carry = INT(result(i:i)) + INT(num1(j:j)) + carry
| 1
Error: ‘a’ argument of ‘int’ intrinsic at (1) must have a numeric type
temp.f95:25:42:
25 | result(i:i) = CHAR(MOD(INT(result(i:i)) + carry, 10) + 48)
| 1
Error: ‘a’ argument of ‘int’ intrinsic at (1) must have a numeric type
temp.f95:26:27:
26 | carry = INT(result(i:i)) + carry
| 1
Error: ‘a’ argument of ‘int’ intrinsic at (1) must have a numeric type
temp.f95:29:42:
29 | result(i:i) = CHAR(MOD(INT(num2(j:j)) + carry, 10) + 48)
| 1
Error: ‘a’ argument of ‘int’ intrinsic at (1) must have a numeric type
temp.f95:30:27:
30 | carry = INT(num2(j:j)) + carry
| 1
Error: ‘a’ argument of ‘int’ intrinsic at (1) must have a numeric type
f951: some warnings being treated as errors
def addStrings(num1: str, num2: str) -> str:
i, j, carry, result = len(num1) - 1, len(num2) - 1, 0, []
while i >= 0 or j >= 0 or carry:
sum = carry
if i >= 0:
sum += int(num1[i])
i -= 1
if j >= 0:
sum += int(num2[j])
j -= 1
carry = sum // 10
result.append(str(sum % 10))
return ''.join(result[::-1])
1. Initialize pointers `i` and `j` to the end of `num1` and `num2`, a `carry` variable to 0, and an empty `result` container.
sum variable with the current carry.
b. Add the current digits from num1[i] and num2[j] to sum if exist, decreasing i and j respectively after each addition.
c. Update the value of carry to be the integer division of sum by 10.
d. Add the value of sum % 10 to the result container.result container as a string.std::string addStrings(std::string num1, std::string num2) {
int i = num1.size() - 1, j = num2.size() - 1, carry = 0;
std::string result;
while (i >= 0 || j >= 0 || carry) {
int sum = carry;
if (i >= 0) {
sum += num1[i] - '0';
--i;
}
if (j >= 0) {
sum += num2[j] - '0';
--j;
}
carry = sum / 10;
result.push_back(sum % 10 + '0');
}
std::reverse(result.begin(), result.end());
return result;
}
1. Initialize pointers `i` and `j` to the end of `num1` and `num2`, a `carry` variable to 0, and an empty `result` container.
sum variable with the current carry.
b. Add the current digits from num1[i] and num2[j] to sum if exist, decreasing i and j respectively after each addition.
c. Update the value of carry to be the integer division of sum by 10.
d. Add the value of sum % 10 to the result container.result container as a string.