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2855:Fibonacci Check-up

题目描述
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.
输入解释
First line is the testcase T.
Following T lines, each line is two integers n, m ( 0<= n <= 10^9, 1 <= m <= 30000 )
输出解释
Output the alpc-number.
输入样例
2
1 30000
2 30000
输出样例
1
3
来自杭电HDUOJ的附加信息
Recommend gaojie

该题目是Virtual Judge题目,来自 杭电HDUOJ

源链接: HDU-2855

最后修改于 2020-10-25T22:57:32+00:00 由爬虫自动更新

共提交 0

通过率 --%
时间上限 内存上限
2000/1000MS(Java/Others) 32768/32768K(Java/Others)