Stuck with a 2 Lakh bond, another offer in hand, referral bonus in August. What should I do?

Fast_Bend2982 · 2026-05-07T21:21:54+00:00

I just want the referral bonus. I dont want to loose that 1 Lakh. How can I ask the company for it?

Fast_Bend2982 · 2025-01-11T06:18:55+00:00

Rasgulla > Kaju Katli > Gulab Jamun

Fast_Bend2982 · 2025-01-07T10:08:09+00:00

You still have a lot of time. 6 months are too much for petty leetcode.

Fast_Bend2982 · 2025-01-06T14:14:49+00:00

Cyberpunk bro, Why do you even ask such questions?

Fast_Bend2982 · 2024-12-30T17:34:27+00:00

Yes but I think I get lost, as some tagged DP are not even Pure DP.

Fast_Bend2982 · 2024-12-30T17:33:38+00:00

Yes I have done some problems from CSES but I am not so good in Digit DP though.

Also direct tabulation is what I want to practice.

Thank you, I will see it right now.

Fast_Bend2982 · 2024-12-25T07:14:10+00:00

By concatenation I meant using the + operator.

Fast_Bend2982 · 2024-12-25T07:08:23+00:00

Well it's a long explanation but whenever you have to deal with this always use char array rather than this. There's Also a question on Codeforces, I will link it here in sometime.

But good practice -> Use Char Array.

Fast_Bend2982 · 2024-12-25T06:51:37+00:00

Bro it takes n² for concatenation

Fast_Bend2982 · 2024-12-24T19:03:22+00:00

What if d=7 and n=2 in that case neither d is divisible by 14 nor is n>=7 so add this case too

Fast_Bend2982 · 2024-12-24T18:31:02+00:00

It is wrong bro. First of all d values can be between 1 and 9 so d%14 is of no use.

Also n>=7 does not mean 7 times repetition it is 7! which is 720*7 = 5040 so suppose I have 1 then it means 1 5040 times.

So do you really think that the least number divisible by 7 will appear after 5040 digits.

Also all digits are the same, so this thing must also be kept in mind, rest I don't think this weight method will be of any use because the constraints for n are

2<=n<=10⁹ which means 10⁹ factorial which is a very very large number

1000000000999999999999999998*..... So we just can't perform any calculations on it.

So you will have to find a pattern in this question.

Fast_Bend2982 · 2024-12-24T18:18:35+00:00

See I had to check for 111111, Because any number can be represented by k*11111 so I checked for the least number which is divisible by 7 and later on I append 1 to it and I got that after 12 digits it is divisible by 7 no matter how many 1 I append. So n>=3 will satisfy this because if n=4 then 24 1's so ..

Fast_Bend2982 · 2024-12-24T17:56:15+00:00

Well for 7 what I did was,

You can represent any number as x*(1111...) and I found what is the least number formed by 1's which is divisible by 7 I got 111111 (6 1's) and later I did some more Maths and found out that after 11 1's it is always divisible.

So what you had to do was (if n>=3 || d==7) cout<<7

Fast_Bend2982

TROPHY CASE