HDU 5950 Recursive sequence（矩阵快速幂）

Problem Description

Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers a and b on a blackboard. And then, the cows would say their identity number one by one. The first cow says the first number a and the second says the second number b. After that, the i-th cow says the sum of twice the (i-2)-th number, the (i-1)-th number, and i4. Now, you need to write a program to calculate the number of the N-th cow in order to check if John’s cows can make it right.

Input

The first line of input contains an integer t, the number of test cases. t test cases follow.
Each case contains only one line with three numbers N, a and b where N,a,b < 231 as described above.

Output

For each test case, output the number of the N-th cow. This number might be very large, so you need to output it modulo 2147493647.

Sample Input

2
3 1 2
4 1 10

Sample Output

85
369
Hint
In the first case, the third number is 85 = 2*1十2十3^4.
In the second case, the third number is 93 = 2*1十1*10十3^4 and the fourth number is 369 = 2 * 10 十 93 十 4^4.

思路

先得到一个递推式

f_{n} = f_{n - 1} + 2 f_{n - 2} + n^{4}

$f_n = f_{n-1} + 2f_{n-2} + n^4$
易得

{\begin{matrix} (n + 1)^{4} \\ F_{n} \\ F_{n - 1} \end{matrix}}

$\left\lbrace\begin{matrix} (n+1)^4\\F_{n}\\F_{n-1} \end{matrix}\right\rbrace$

=

$=$

{\begin{matrix} (\frac{n + 1}{n})^{4} & 0 & 0 \\ 1 & 1 & 2 \\ 0 & 1 & 0 \end{matrix}}

$\left\lbrace\begin{matrix} (\frac{n+1}{n})^4 & 0 & 0\\ 1 & 1 & 2 \\ 0 & 1 & 0 \end{matrix}\right\rbrace$

\cdot

$·$

{\begin{matrix} n^{4} \\ F_{n - 1} \\ F_{n - 2} \end{matrix}}

$\left\lbrace\begin{matrix} n^4\\F_{n-1}\\F_{n-2} \end{matrix}\right\rbrace$
然而系数矩阵与n相关不能使用
所以我们需要使系数矩阵变的与n无关
系数矩阵要与n无关，需要对

n^{4}

$n^4$ 进行二项式展开

F_{n} = F_{n - 1} + 2 F_{n - 2} + n^{4}

$F_{n} = F_{n-1} + 2F_{n-2} + n^4$

= F_{n - 1} + 2 F_{n - 2} + (n + 1 - 1)^{4}

$=F_{n-1} + 2F_{n-2} + (n+1-1)^4$

= F_{n - 1} + 2 F_{n - 2} + \sum_{i = 1}^{n} C_{n}^{i} (n - 1)^{i}

$= F_{n-1} + 2F_{n-2} +\sum\limits_{i=1}^n{C_n^i(n-1)^i}$

{\begin{matrix} F_{n} \\ F_{n - 1} \\ n^{0} \\ n^{1} \\ n^{2} \\ n^{3} \\ n^{4} \end{matrix}}

$\left\lbrace\begin{matrix} F_n\\F_{n-1}\\n^0\\n^1\\n^2\\n^3\\n^4 \end{matrix}\right\rbrace$

=

$=$

{\begin{matrix} 1 & 2 & 1 & 4 & 6 & 4 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 1 & 3 & 3 & 1 & 0 \\ 0 & 0 & 1 & 4 & 6 & 4 & 1 \end{matrix}}

$\left\lbrace\begin{matrix} 1&2&1&4&6&4&1\\ 1&0&0&0&0&0&0\\ 0&0&1&0&0&0&0\\ 0&0&1&1&0&0&0\\0&0&1&2&1&0&0\\ 0&0&1&3&3&1&0\\ 0&0&1&4&6&4&1 \end{matrix}\right\rbrace$

\cdot

$·$

{\begin{matrix} F_{n - 1} \\ F_{n - 2} \\ (n - 1)^{0} \\ (n - 1)^{1} \\ (n - 1)^{2} \\ (n - 1)^{3} \\ (n - 1)^{4} \end{matrix}}

$\left\lbrace\begin{matrix} F_{n-1}\\F_{n-2}\\(n-1)^0\\(n-1)^1\\(n-1)^2\\(n-1)^3\\(n-1)^4 \end{matrix}\right\rbrace$

可以看下这篇博客学习构造系数矩阵根据递推公式构造系数矩阵用于快速幂
的

代码

#include<stdio.h>
#include<string.h>
using namespace std;

#define ll long long
const ll maxn = 7;
const ll mod = 2147493647;

struct mat
{
    ll m[maxn][maxn];
};

mat operator *(mat x, mat y)
{
    mat ret;
    for(int i = 0; i < maxn; i++)
    {
        for(int j = 0; j < maxn; j++)
        {
            ret.m[i][j] = 0;
            for(int k = 0; k < maxn; k++)
            {
                ret.m[i][j] = (x.m[i][k] * y.m[k][j] + ret.m[i][j]) % mod;
            }
        }
    }
    return ret;
}

mat pow_mat(mat a, ll fuck)
{
    mat ret;
    memset(ret.m,0,sizeof(ret.m));
    for(int i = 0; i < maxn; i++) ret.m[i][i] = 1;
    while(fuck)
    {
        if(fuck&1) ret = ret*a;
        a = a*a;
        fuck >>= 1;
    }
    return ret;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll f1, f2, m;
        scanf("%lld%lld%lld",&m,&f1,&f2);
        if(m == 1) printf("%lld\n",f1);
        else if(m == 2) printf("%lld\n",f2);
        else
        {
            mat a;
            memset(a.m,0,sizeof(a.m));
            a.m[0][0] = f2;
            a.m[1][0] = f1;
            a.m[2][0] = 1;
            a.m[3][0] = 2;
            a.m[4][0] = 4;
            a.m[5][0] = 8;
            a.m[6][0] = 16;
            mat b =
            {
                1, 2, 1, 4, 6, 4, 1,
                1, 0, 0, 0, 0, 0, 0,
                0, 0, 1, 0, 0, 0, 0,
                0, 0, 1, 1, 0, 0, 0,
                0, 0, 1, 2, 1, 0, 0,
                0, 0, 1, 3, 3, 1, 0,
                0, 0, 1, 4, 6, 4, 1
            };
            a = pow_mat(b,m-2)*a;
            printf("%lld\n",a.m[0][0]);
        }
    }
    return 0;
}