矩阵与斐波那契数列

Fibonacci Check-up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1436    Accepted Submission(s): 825

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

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
const int N=2;
struct Matrix{
    LL m[N][N];
};
Matrix A={
    1,1,
    1,0
};
Matrix I={
    1,0,
    0,1
};
Matrix multi(Matrix a,Matrix b,int mod)
{
    Matrix c;
    for(int i=0;i<N;i++)
    {
        for (int j=0;j<N;j++)
        {
            c.m[i][j]=0;
            for(int k=0;k<N;k++)
            {
                c.m[i][j]+=a.m[i][k]*b.m[k][j]%mod;
            }
            c.m[i][j]%=mod;
        }
    }
    return c;
}
Matrix power(Matrix A,int k,int mod)
{
    Matrix ans=I,p=A;
    while(k)
    {
        if(k&1)
        {
            ans=multi(ans,p,mod);
            k--;
        }
        k>>=1;
        p=multi(p,p,mod);
    }
    return ans;
}
int main()
{
    int n,mod;
    LL T;
    cin>>T;
    while(T--)
    {
        scanf("%d%d",&n,&mod);
        if(n==0)
            printf("0\n");
        else{
            n*=2;
        Matrix ans=power(A,n-1,mod);
        printf("%I64d\n",ans.m[0][0]);
        }
    }
    return 0;
}