HDU 2855:Fibonacci Check-up_斐波拉契问题

Fibonacci Check-up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 672    Accepted Submission(s): 370

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

  
  

   
   2
1 30000
2 30000
  
  

 

Sample Output

  
  

   
   1
3
  
  

 

Source

2009 Multi-University Training Contest 5 - Host by NUDT

 

Recommend

gaojie

 

//09年的一场多校联合赛的题目，推公式推了一天才搞定！设ｓ[n]=sigma(C(n,k)*F(k)),打表之后发现规律S[n]=3*S[n-1]-S[n-2]，这样其实就可以构造矩阵乘法解决了，但是这样的规律毕竟只是猜测的，需要证明出来着实不会！假想此公式成立，对于这样的等式构造二介特征方程解出来之后S[n]=1/sqrt(5)*((（3+sqrt(5)）/2)^n - ((3-sqrt(5))/2)^5 ).   对于((3+sqrt(5)/2)^n=(6+2*sqrt(5)/4)^n= (1+sqrt(5)^2/2^2)^n=(1+sqrt(5)/2)^(2*n) 。所以可得S[n]=F[2*n]，当然这只是猜想，但是由刚才的推导可以猜想如果把F[n]=1/sqrt(5)*(((1+sqrt(5))/2)^n-((1-sqrt(5))/2)^n)带入ｓ[n]=sigma(C(n,k)*F(k))推导之后等式会怎样，果然得到F[2*n]，所以本题其实就是让求F[2*n]。

#include<iostream>
using namespace std;
#define LL long long

LL fib_last(LL n,LL m)
{
	if(n==0)	return 0;
	LL ret=1,next=1,ret_=ret;
	LL flag=1,tt=n;
	while(tt>>=1)	flag<<=1;
	while(flag>>=1)
	{
		if(n&flag)
		{
			ret_=ret*ret+next*next;
			next=(ret+ret+next)*next;
		}
		else
		{
			ret_=(next+next+m-ret)*ret;
			next=ret*ret+next*next;
		}
		ret=ret_%m;
		next=next%m;
	}
	return ret;
}

int main()
{
	int t;
	LL n,m;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%I64d%I64d",&n,&m);
		printf("%I64d\n",fib_last(2*n,m)%m);
	}
	return 0;
}