HDU 1757 A Simple Math Problem 矩阵快速幂

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);

对于这样一个式子，通过矩阵与线性变换的关系，可以轻松的构造出这样的矩阵

A0:

9
8
7
6
5
4
3
2
1
0

A1:
1 1 1 1 1 1 1 1 1 1
1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0

那么f(n)=A1^(n-9)*A0

快速幂即可

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <climits>
#include <iostream>
#include <string>

using namespace std;
 
#define MP make_pair
#define PB push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF = INT_MAX / 3;
const double eps = 1e-8;
const LL LINF = 1e17;
const double DINF = 1e60;
const int maxn = 11;
LL k, mod;

struct Matrix {
    int n, m;
    LL data[maxn][maxn];
    Matrix(int n = 0, int m = 0): n(n), m(m) {
        memset(data, 0, sizeof(data));
    }
};

Matrix operator * (Matrix a, Matrix b) {
    Matrix ret(a.n, b.m);
    for(int i = 1; i <= a.n; i++) {
        for(int j = 1; j <= b.m; j++) {
            for(int k = 1; k <= a.m; k++) {
                ret.data[i][j] += a.data[i][k] * b.data[k][j];
                ret.data[i][j] %= mod;
            }
        }
    }
    return ret;
}

Matrix pow(Matrix mat, LL p) {
    if(p == 1) return mat;
    Matrix ret = pow(mat * mat, p / 2);
    if(p & 1) ret = ret * mat;
    return ret;
}

int main() {
    int a[10];
    while(cin >> k >> mod) {
        for(int i = 0; i < 10; i++) cin >> a[i];
        Matrix A(10, 10), A0(10, 1);
        if(k < 10) cout << k % mod << endl;
        else {
            for(int i = 1; i <= 10; i++) A.data[1][i] = a[i - 1];
            for(int i = 2; i <= 10; i++) A.data[i][i - 1] = 1;
            for(int i = 1, j = 9; i <= 10; i++, j--) A0.data[i][1] = j;
            A = pow(A, k - 9);
            A0 = A * A0;
            cout << A0.data[1][1] << endl;
        }
    }
    return 0;
}