【HDU 2855】 Fibonacci Check-up （矩阵乘法）

Fibonacci Check-up

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

【题意】

　　求S(n)=∑C[k][n]*Fibonacci(k) mod m(0<=k<=n)
　　( 0<= n <= 10^9, 1 <= m <= 30000 )

【分析】

　　组合数和斐波那契数列都是很有特点的东西，然而我想了一会儿还是没有想出来。

　　现在又懂得了一点，能写出递推式，像斐波那契数列一样的，它的第k项其实可以表示成矩阵的幂，即A^k，把它当成数一样考虑就很方便。

　　对于组合数，二项式定理啊真是太厉害了。。终于有点懂母函数的思想啊....

　　

图片转自：http://blog.csdn.net/hzh_0000/article/details/38171903

其实还有第二种方法，我没打，感觉我不太可能推出来。。

第一种方法代码如下：

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<queue>
 #include<cmath>
 using namespace std;

 struct node
 {
     int a[][];
 }t[];

 int n,m;

 void init()
 {
     t[].a[][]=;t[].a[][]=;
     t[].a[][]=;t[].a[][]=;
 }

 void mul(int x,int y,int z)
 {
     for(int i=;i<=;i++)
      for(int j=;j<=;j++)
      {
          t[].a[i][j]=;
          for(int k=;k<=;k++)
             t[].a[i][j]=(t[].a[i][j]+t[y].a[i][k]*t[z].a[k][j])%m;
      }
     t[x]=t[];
 }

 void get_un()
 {
     memset(t[].a,,sizeof(t[].a));
     for(int i=;i<=;i++) t[].a[i][i]=;
 }

 void qpow(int b)
 {
     get_un();
     while(b)
     {
         if(b&) mul(,,);
         mul(,,);
         b>>=;
     }
 }

 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&m);
         init();
         qpow(n);
         printf("%d\n",t[].a[][]);
     }
     return ;
 }

[HDU 2855]

2016-09-28 14:10:22