I’m getting WA in the second subtask in BUGCAL… Why?
My solution is in O(number of digits). I don’t know why it is giving TLE. This is my solution.
It is in Python. My logic is I calculate the number if I remove one digit, and check if it is divisible by 6 and greater than previously held value and replace it. Then append zeroes according to input if the first digit is removed.
e.g:- 3268398 is input,
num will hold the values (268398, 368398, 328398, 326398, 326898, 326838, 326839) in each iteration respectively.
Will check if it is divisible by 6 and greater than previously divisible value.
Thanks in advance.
I don’t know why i am getting RE(NZEC)?
this is my solution https://www.codechef.com/viewsolution/15995869
My code gives wrong ans.Any test case where it may fail?
link:https://www.codechef.com/viewsolution/16002121
for the second question ,every test case i checked is satisfied still i am getting a wrong answer please help me out pointing out just 1 test case that this code cannot run ??
https://www.codechef.com/viewsolution/15988759
What is wrong in this code for BUGCAL? PLease help. Everything handled!
Taran, Can you help me debug my code? It seems like I’m missing some corner cases.
Edit: I’m sorry. The link was wrong. This is link to solution. https://www.codechef.com/viewsolution/16002813
Hey @shivam2296 ur code showing wrong for a.length<b.length
Example:
input:
2
11 1145
12 13456
output:
-)
)9
Hey taran I am getting wa for the 2nd subtask in the problem BUGCAL. Any testcase where I might fail. Here’s the link to my solution - link text
you are getting -1 for 66666666666
Your Code gives wrong input for
1
99999999 99999999
output:
88888880
But Unable to figure out the exact reason for WA
@taran_1407 @vijju check out this code why showing wrong
it’s showing Acess Denied
Thats because of pow10 function, which stores result in double. Due to precision loss of ~{10}^{-8}, it shows deviation at large values.
Ur code printing leading Zeroes!!
Mate, you are complicating your life a lot by using that vector V. Simply check for digit remainder 3 during a loop, check if sum%3 == d%3. It will give same reaults.
Refer to my implementation and feep free to ask anythibg…
A must know thing for python coders. When dealing with large number, Even basic arithmetic operations like remainder cost time proportional to number of digits, making your solution O(N^2).
Feel free to ask anything…
Mate, you cannot take a number greater than or equal to 2^31 or smaller than -2^31 as int in java.
Take input as
String p = obj.next();
Feel free to ask anything…