For two strings s and t, we say "t divides s " if and only if s = t + ... + t (i.e., t is concatenated with itself one or more times).
Given two strings str1 and str2, return the largest string x such that x divides both str1 and str2.
Example 1:
Input: str1 = "ABCABC ", str2 = "ABC " Output: "ABC "
Example 2:
Input: str1 = "ABABAB ", str2 = "ABAB " Output: "AB "
Example 3:
Input: str1 = "LEET ", str2 = "CODE " Output: " "
Constraints:
1 <= str1.length, str2.length <= 1000str1 and str2 consist of English uppercase letters.program main
implicit none
character(len=:), allocatable :: str1, str2, x
integer :: i
! Example 1
str1 = "ABCABC "
str2 = "ABC "
call solve(str1, str2, x)
write (*, '(A)') trim(x)
! Example 2
str1 = "ABABAB "
str2 = "ABAB "
call solve(str1, str2, x)
write (*, '(A)') trim(x)
! Example 3
str1 = "LEET "
str2 = "CODE "
call solve(str1, str2, x)
write (*, '(A)') trim(x)
contains
subroutine solve(str1, str2, x)
implicit none
character(len=*), intent(in) :: str1, str2
character(len=:), allocatable, intent(out) :: x
integer :: n
n = len(str2)
do while (n > 0)
if (index(str1, str2) /= 0 .and. index(str2, str2) == 0) then
x = str2
exit
end if
n = n - 1
str2 = str2 // str2
end do
end subroutine solve
end program maintemp.f95:40:12:
40 | str2 = str2 // str2
| 1
Error: Dummy argument ‘str2’ with INTENT(IN) in variable definition context (assignment) at (1)
program longest_common_divisor
implicit none
character(len=1000) :: str1, str2, x
integer :: i, j, k, l, m, n
read(*,*) str1, str2
! Initialize x as an empty string
x = ""
! Loop through all possible lengths of x
do i = 1, len(str1)
! Loop through all possible starting indices of x in str1
do j = 1, len(str1) - i + 1
! Loop through all possible lengths of x in str2
do k = 1, len(str2)
! Loop through all possible starting indices of x in str2
do l = 1, len(str2) - k + 1
! Check if x divides both str1 and str2
if (str1(j:j+i-1) == str2(l:l+k-1)) then
x = str1(j:j+i-1)
exit
end if
end do
if (x /= "") exit
end do
if (x /= "") exit
end do
if (x /= "") exit
end do
! Output the largest string x that divides both str1 and str2
write(*,*) x
end program longest_common_divisor
This script uses a brute force approach to solve the problem. It loops through all possible lengths of the string `x` and checks if `x` divides both `str1` and `str2`. If `x` divides both strings, it is the largest string that divides both strings. The script uses the `len` function to get the length of the strings and the `==` operator to check if two strings are equal. The `exit` statement is used to exit the loops early if `x` is found to divide both strings.
Note that this script assumes that the input strings are uppercase and contain only English uppercase letters. If the input strings can contain other characters, you may need to modify the script accordingly.
temp.f95:49:132:
49 | This script uses a brute force approach to solve the problem. It loops through all possible lengths of the string `x` and checks if `x` divides both `str1` and `str2`. If `x` divides both strings, it is the largest string that divides both strings. The script uses the `len` function to get the length of the strings and the `==` operator to check if two strings are equal. The `exit` statement is used to exit the loops early if `x` is found to divide both strings.
| 1
Error: Line truncated at (1) [-Werror=line-truncation]
temp.f95:51:132:
51 | Note that this script assumes that the input strings are uppercase and contain only English uppercase letters. If the input strings can contain other characters, you may need to modify the script accordingly.
| 1
Error: Line truncated at (1) [-Werror=line-truncation]
f951: some warnings being treated as errors
def prefixesDivBy5(nums):
result = []
current = 0
for num in nums:
current = ((current << 1) | num) % 5
result.append(current == 0)
return result
The problem requires to find out if the binary representation of the subarray from 0 to i is divisible by 5. To solve this, we can maintain an integer current that keeps track of the value of this binary representation modulo 5.
Iterate through the nums array, and, for each number in the array, update the running total current as follows:
- Shift current one bit to the left (multiply by 2) using the bitwise left shift operator << 1
- Perform bitwise OR operation with the current number num to add it as the least significant bit to the binary representation
- Compute this binary representation modulo 5
After computing the current binary representation (modulo 5), check if it is divisible by 5. If it's divisible, then the value of current would be 0. Append the result (true or false) to the answer list or vector.
Finally, return the resultant list of booleans.###
End ###
#include <vector>
std::vector<bool> prefixesDivBy5(std::vector<int>& nums) {
std::vector<bool> result;
int current = 0;
for (int num : nums) {
current = ((current << 1) | num) % 5;
result.push_back(current == 0);
}
return result;
}
The problem requires to find out if the binary representation of the subarray from 0 to i is divisible by 5. To solve this, we can maintain an integer current that keeps track of the value of this binary representation modulo 5.
Iterate through the nums array, and, for each number in the array, update the running total current as follows:
- Shift current one bit to the left (multiply by 2) using the bitwise left shift operator << 1
- Perform bitwise OR operation with the current number num to add it as the least significant bit to the binary representation
- Compute this binary representation modulo 5
After computing the current binary representation (modulo 5), check if it is divisible by 5. If it's divisible, then the value of current would be 0. Append the result (true or false) to the answer list or vector.
Finally, return the resultant list of booleans.###
End ###