The Ultra Fast Mathematician problem given in bitmasking section but the editorial is in string approach can i get the editorial with bit masking ans the output format is written as we should not omit the leading zeros but when i unlocked the editorial the output where in leading zero.
Here is my solution https://ideone.com/NmV4QA with bit masking but cant able to add leading zero
Can i get a solution with bit masking
Hello @Axelblaze1005 in this question we have to just perform the xor of the bits which says if the bits are same then the output should be low whereas if the bits are different then the result will be high(i.e 1) so as we are already given with the bits so thats why we are taking the bits in the string format and then just checking of they are same or different…
wait i have checed your code and you are also taking the if condition for the xor of the bits.
looking for error in your code.
please wait.
setting the jth bit as 1
Hey @Axelblaze1005 you cant do this question with the help of the bits,you have to do this with the string because the cinstraints are high.
even if you are using long long int then you cant take the work on the number with digits more than 18.
and here the maximum number of digits can be 100.
i am asking ans = ans|(1<<j);?
what is this | in the line.
i think this is wrong syntax.
| is using for or like we used ^ for xor
okh thanks for helping
or symbol is used with this ‘||’.
i am not checking the condition i am setting the bit with or like this
ohh okay okay.do you have anyother doubt?
or is your doubt cleared?
no other doubts thanks for helping
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