CRZYSMKR - Crazy Smoker
The "BHAI Group" Of IIIT Allahabad is Famous For Many Things,Leading In Every Field Of College Activity
So One Day The Leader Of Bhai Group decided to smoke C(N) cigarettes each day:
- F(N) = 34^N+ (30 x N) + 32
- C(N) = F(N) mod (11), where x mod (y) is the remainder obtained by diving x by y.
But Bhai Group's Leader's Girlfriend wants that he doesn't smoke any cigarette, so she made modifications:
- F(N) = 34^N+ (30 x N) + (32 + M)
- C(N) = F(N) mod (11)
Edit 1 : Time Limit Set To .100s
Problem Credits : IIIT Allahabad HE Club
First line of each test case is an integer T, total number of test cases. Next T lines contains a single integer N.
Print the minimum value of M in single line for each test case.
1 <= T <= 10^6
1 <= N <= 10^18
Input:2 1 2Output:3 6Explaination :
For N = 1 F(N) = 34 + 30 + 32 = 96 So, M = 3 Now, C(N) = 99 mod(11) = 0 For N = 2 F(N) = 1156 + 60 + 32 = 1248 So, M = 6 Now, C(N) = 1254 mod(11) = 0
@golu20174024 thanks for sharing but we dont really need to know
I just love penis!Last edit: 2020-11-01 21:05:08
No need to calculate 34^N, because 34^N % 11 is always 1 and (a+b+c)%N = a%N +b%N +c%N
Don't believe spojtoolkit for this one.
Super easy, just follow modular arithmetic closely!!Last edit: 2017-12-05 13:15:05
AC in a GO!!
those who know binomial expansion can do it in easily !
34^(10^18) is beyond C limits, costed me 2 WA.
Let me simplify the stupid problem statement :