【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[5][5];
}t[5];

int n,m;

void init()
{
    t[0].a[1][1]=1;t[0].a[1][2]=1;
    t[0].a[2][1]=1;t[0].a[2][2]=2;
}

void mul(int x,int y,int z)
{
    for(int i=1;i<=2;i++)
     for(int j=1;j<=2;j++)
     {
         t[2].a[i][j]=0;
         for(int k=1;k<=2;k++)
            t[2].a[i][j]=(t[2].a[i][j]+t[y].a[i][k]*t[z].a[k][j])%m;
     }
    t[x]=t[2];
}

void get_un()
{
    memset(t[1].a,0,sizeof(t[1].a));
    for(int i=1;i<=2;i++) t[1].a[i][i]=1;
}

void qpow(int b)
{
    get_un();
    while(b)
    {
        if(b&1) mul(1,0,1);
        mul(0,0,0);
        b>>=1;
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        init();
        qpow(n);
        printf("%d\n",t[1].a[1][2]);
    }
    return 0;
}

 

2016-09-28 14:10:22