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

21 300002 30000

Sample Output

13

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 <stdio.h>#include <string.h>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;}