HDU 1757 A Simple Math Problem [矩阵快速幂]

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题目链接： HDU 1757 A Simple Math Problem

Problem Description

Lele now is thinking about a simple function f(x).

If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .

Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.

Input

The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.

Output

For each case, output f(k) % m in one line.

Sample Input

10 9999
1 1 1 1 1 1 1 1 1 1
20 500
1 0 1 0 1 0 1 0 1 0

Sample Output

45
104

题解：

矩阵的快速幂
容易发现

$f (x) = [f (x - 1) f (x - 2) \dots f (x - 10)] ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ a 0 a 1 ⋮ a 9 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥$
要用快速幂，就要凑成方阵的形式，观察发现Mi∗A=Mi+1 的形式：
$⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ f (x - 1) f (x - 2) ⋮ f (x - 10) f (x - 2) f (x - 3) ⋮ \dots \dots \dots ⋱ \dots f (x - 10) f (x - 11) ⋮ f (x - 19) ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ a 0 a 1 a 3 ⋮ a 9 100 ⋮ 0 010 ⋮ 0 001 ⋮ 0 \dots \dots \dots ⋱ \dots 0 ⋮ ⋮ 10 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ f (x) f (x - 1) ⋮ f (x - 9) f (x - 1) f (x - 2) ⋮ \dots \dots \dots ⋱ \dots f (x - 9) f (x - 10) ⋮ f (x - 18) ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥$
初始矩阵M为
$⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ f (19) f (18) ⋮ f (10) f (18) f (17) ⋮ \dots \dots \dots ⋱ \dots f (10) f (9) ⋮ f (1) ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥$
x 小于20，可以用公式直接求，反正数据小，没有优化的必要。x 大于20, 只需要用初始矩阵乘以A矩阵
$⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ a 0 a 1 a 3 ⋮ a 9 100 ⋮ 0 010 ⋮ 0 001 ⋮ 0 \dots \dots \dots ⋱ \dots 0 ⋮ ⋮ 10 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥$
的 x−19 次方
这里就可以用到快速幂来求解。

代码：

#include <iostream>#include <cstring>using namespace std;typedef int Mat[11][11];int k, m;Mat A, T, M;int a[20];void matmul(Mat a, Mat b) {    Mat c;    memset(c, 0, sizeof(c));    for (int i = 0; i < 10; i++) {        for (int j = 0; j < 10; j++) {            for (int k = 0; k < 10; k++) {                c[i][j] += a[i][k]*b[k][j];            }            c[i][j] %= m;        }    }    for (int i = 0; i < 10; i++)         for (int j = 0; j < 10; j++)            a[i][j] = c[i][j];}int fun(int x) {    if (x < 10) return x%m;    else {        int sum = 0;        for (int i = 0; i < 10; i++) {            if (a[i]==1) {                sum += fun(x-(i+1));            }        }        return sum%m;    }}int main() {    while (cin >> k >> m) {        memset(A, 0, sizeof(A));        memset(M, 0, sizeof(M));        memset(T, 0, sizeof(T));        for (int i = 0; i < 9; i++) {            T[i][i+1] = 1;        }        for (int i = 0; i < 10; i++) {            cin >> a[i];            T[i][0] = a[i];            A[i][i] = 1;        }        if (k < 20) {            cout << fun(k)%m << endl;        }        else {            k -= 19;            for (int i = 0; i < 10; i++) {                for (int j = 0; j < 10; j++) {                    M[i][j] = fun(20-i-j-1);                }            }            while (k > 0) {                if (k&1) matmul(A, T);                k >>= 1;                matmul(T,T);            }            matmul(M, A);            cout << M[0][0] << endl;        }       }    return 0;}

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