HDU 3306 Another kind of Fibonacci ---构造矩阵***

Another kind of Fibonacci

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1219    Accepted Submission(s): 466

Problem Description

As we all known , the Fibonacci series : F(0) = 1, F(1) = 1, F(N) = F(N - 1) + F(N - 2) (N >= 2).Now we define another kind of Fibonacci : A(0) = 1 , A(1) = 1 , A(N) = X * A(N - 1) + Y * A(N - 2) (N >= 2).And we want to Calculate S(N) , S(N) = A(0)² +A(1)²+……+A(n)².

 

Input

There are several test cases.
Each test case will contain three integers , N, X , Y .
N : 2<= N <= 2³¹ – 1
X : 2<= X <= 2³¹– 1
Y : 2<= Y <= 2³¹ – 1

 

Output

For each test case , output the answer of S(n).If the answer is too big , divide it by 10007 and give me the reminder.

 

Sample Input

2 1 1

3 2 3

 

Sample Output

6

196

 

Author

wyb

 

同学说，矩阵这一块，最难到如何构造矩阵，这题是构造矩阵的经典例题。

 

如果构造的呢？？

A(N)= X*A(N-1)  +  Y*A(N-2)

S(N)= S(N-1)      +  A(N)^2;

合并一下

S(N)= S(N-1) + X^2*A(N-1)^2 + Y^2*A(N-2) +2*X*Y*A(N-1)A(N-2);

 

很像做过到一道题目：HDU 1757  f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);

那么把参数提取出来就有4个了:

                        S(N-1)     A(N-1)^2   A(N-2)%^2     A(N-1)A(N-2)

对应到系数         1               X^2           Y^2                 2*X*Y

 

| 1     X^2      Y^2       2*X*Y  |   |  S(N-1)           |

| 0     X^2      Y^2       2*X*Y  |   |  A(N-1)^2       |

| 0        1          0            0       |  |  A(N-2)^2       |

| 0       X           0            Y       |  |  A(N-1)A(N-2) |

 

 

自己推一推就可以的。

这一题还有一个地方需要注意：

X : 2<= X <= 2³¹– 1
Y : 2<= Y <= 2³¹ – 1

注意相乘时的溢出。

 

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

 struct node
 {
     __int64 mat[][];
 }M_hxl,M_tom;

 void make_init(__int64 x,__int64 y)
 {
     M_hxl.mat[][]=;
     M_hxl.mat[][]=(x*x)%;
     M_hxl.mat[][]=(y*y)%;
     M_hxl.mat[][]=(*x*y)%;

     M_hxl.mat[][]=;
     M_hxl.mat[][]=(x*x)%;
     M_hxl.mat[][]=(y*y)%;
     M_hxl.mat[][]=(*x*y)%;

     M_hxl.mat[][]=;
     M_hxl.mat[][]=;
     M_hxl.mat[][]=;
     M_hxl.mat[][]=;

     M_hxl.mat[][]=;
     M_hxl.mat[][]=x;
     M_hxl.mat[][]=;
     M_hxl.mat[][]=y;
 }

 void make_first(node *cur)
 {
     __int64 i,j;
     for(i=;i<=;i++)
     for(j=;j<=;j++)
     if(i==j)
     cur->mat[i][j]=;
     else cur->mat[i][j]=;
 }

 struct node cheng(node cur,node now)
 {
     node ww;
     __int64 i,j,k;
     memset(ww.mat,,sizeof(ww.mat));
     for(i=;i<=;i++)
     for(k=;k<=;k++)
     if(cur.mat[i][k])
     {
         for(j=;j<=;j++)
         if(now.mat[k][j])
         {
             ww.mat[i][j]+=cur.mat[i][k]*now.mat[k][j];
             if(ww.mat[i][j]>=)
             ww.mat[i][j]%=;
         }
     }
     return ww;
 }
 void power_sum2(__int64 n)
 {
     __int64 sum=;
     make_first(&M_tom);
     while(n)
     {
         if(n&)
         {
             M_tom=cheng(M_tom,M_hxl);
         }
         n=n>>;
         M_hxl=cheng(M_hxl,M_hxl);
     }
     sum=sum+*M_tom.mat[][]+M_tom.mat[][]+M_tom.mat[][]+M_tom.mat[][];
     if(sum>=)
     sum=sum%;
     printf("%I64d\n",sum);

 }

 int main()
 {
     __int64 n,x,y;
     while(scanf("%I64d%I64d%I64d",&n,&x,&y)>)
     {
         x=x%;//防止溢出
         y=y%;//防止溢出
         memset(M_hxl.mat,,sizeof(M_hxl.mat));
         make_init(x,y);
         power_sum2(n-);
     }
     return ;
 }