hdu-2855 Fibonacci Check-up

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

解题思路：
和的形式为二项式展开，故所求为(A+E)^n,E为单位矩阵，然后快速幂；

#include 
#include 

struct Matrix{
    int arr[2][2];
};
int n,mod;
Matrix Mul(Matrix a,Matrix b) 
{
    Matrix c;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
        {
            c.arr[i][j]=0;
            for(int k=0;k<2;k++)
                c.arr[i][j]=(c.arr[i][j]+a.arr[i][k]*b.arr[k][j]%mod)%mod;
            c.arr[i][j]%mod;
        }
    return c;
}
Matrix Pow(Matrix a,int m)
{
    Matrix b;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            if(i==j)
                b.arr[i][j]=1;
            else
                b.arr[i][j]=0; 
    while(m) 
    {
        if(m&1==1)
            b=Mul(b,a);
        a=Mul(a,a);
        m>>=1;
    }
    return b;
}
int main ()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&mod);
        Matrix orig;
        for(int i=0;i<2;i++)
            for(int j=0;j<2;j++)
                orig.arr[i][j]=1;
        orig.arr[0][0]=2;
        Matrix ans;
        ans=Pow(orig,n);
        printf("%d\n",ans.arr[0][1]%mod);
    }
    return 0;
}