HDU(2855_矩阵的二分快速幂)

Fibonacci Check-up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 900    Accepted Submission(s): 507

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

21 300002 30000

 

Sample Output

13

 

Source

2009 Multi-University Training Contest 5 - Host by NUDT

 

Recommend

gaojie

这个如果能够推出答案当然就能求出数据，

但是难就难在你推不出来。。。

诶，这是真的有点困难。

贴出别人的推导过程：

贴出自己的代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <string>using namespace std;int N, M;struct Matrix{int date[2][2];void setE(){for (int i = 0; i < 2; i++){for (int j = 0; j < 2; j++){date[i][j] = (i == j);}}}Matrix Mul(Matrix m){Matrix e;for (int i = 0; i < 2; i++){for (int j = 0; j < 2; j++){e.date[i][j] = 0;for (int k = 0; k < 2; k++){e.date[i][j] += (long long int)date[i][k] * m.date[k][j] % M;}e.date[i][j] %= M;}}return e;}}mat;Matrix quick_pow(Matrix m, int n){Matrix ans;ans.setE();while (n){if (n & 1){ans = ans.Mul(m);}m = m.Mul(m);n >>= 1;}return ans;}int main(){int T;while (scanf("%d", &T) != EOF){while (T--){scanf("%d%d", &N, &M);if (N == 0){printf("0\n");continue;}mat.date[0][0] = 0;mat.date[0][1] = 1;mat.date[1][0] = 1;mat.date[1][1] = 1;Matrix m = quick_pow(mat, 2 * N - 1);printf("%d\n", (m.date[0][0] + m.date[1][0]) % M);}}//system("pause");return 0;}