hdu 2855 Fibonacci Check-up 矩阵快速幂&&模板

Fibonacci Check-up

 HDU - 2855 

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

从s[0] 递推一下

可以发现 s[i] =f[2i]

这样通过矩阵快速幂求斐波那契数列就行了 

1 1

1 0 是初始矩阵

剩下的套个矩阵快速幂的模板

[code]#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

const int MAX_N = 3;

int N, MOD;

void multipy( int a[MAX_N][MAX_N], int b[MAX_N][MAX_N], int c[MAX_N][MAX_N] ){
for( int i = 1; i <= 2; i++ ){
for( int j = 1; j <= 2; j++ ){
c[i][j] = 0;
for( int k = 1; k <= 2; k++ ){
c[i][j] = ( c[i][j] + a[i][k] * b[k][j] ) % MOD;
}
}
}
}

void get_matrix_pow( int a[MAX_N][MAX_N], int n ){
int ans[MAX_N][MAX_N] = {0};
int temp[MAX_N][MAX_N];
for( int i = 1; i <= 2; i++ )   ans[i][i] = 1;

while( n ){
if( n % 2 == 1 ){
multipy( ans, a, temp );
memcpy( ans, temp, sizeof( int ) * MAX_N * MAX_N );
}
multipy( a, a, temp );
memcpy( a, temp, sizeof( int ) * MAX_N * MAX_N );
n /= 2;
}
memcpy( a, ans, sizeof( int ) * MAX_N * MAX_N );
}

int main(){
int T;

scanf( "%d", &T );
while( T-- ){
scanf( "%d%d", &N, &MOD );
if( N == 0 ){
printf( "%d\n", 0 );
continue;
}else{
int a[MAX_N][MAX_N];
a[1][1] = 1;a[1][2] = 1;
a[2][1] = 1;a[2][2] = 0;
get_matrix_pow( a, 2 * N );
printf( "%d\n", a[2][1] );
}
}
return 0;
}