Fibonacci Check-up
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 615Accepted Submission(s): 333
Problem Description
Every ALPC has his own alpc-number just like alpc12, alpc55, alpc62 etc.
As more and more fresh man join us. How to number them? And how to avoid their alpc-number conflicted?
Of course, we can number them one by one, but that’s too bored! So ALPCs use another method called Fibonacci Check-up in spite of collision.
First you should multiply all digit of your studying number to get a number n (maybe huge).
Then use Fibonacci Check-up!
Fibonacci sequence is well-known to everyone. People define Fibonacci sequence as follows: F(0) = 0, F(1) = 1. F(n) = F(n-1) + F(n-2), n>=2. It’s easy for us to calculate F(n) mod m.
But in this method we make the problem has more challenge. We calculate the formula
, is the combination number. The answer mod m (the total number of alpc team members) is just your alpc-number.
Input
First line is the testcase T.
Following T lines, each line is two integers n, m ( 0<= n <= 10^9, 1 <= m <= 30000 )
Output
Output the alpc-number.
Sample Input
Sample Output
Source
有一个公式:
C(N,1)F1+C(N,2)F2+C(N,3)F3+...+C(N,N)FN=F(2N)
证明方法我有点忘了,大致是利用那个比内公式去推
n=0和1最好特判一下
我的代码:
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