HDOJ 1757 – A Simple Math Problem

Matrix Multiplication (& Quick Power) 

Description

给出一个递推式，求f(x) 对 m 取模的结果。

Type

Matrix Multiplication

Quick Power

Analysis

很典型的矩阵乘法和快速幂的题目，需要构造 10 * 10 的矩阵。矩阵也是很经典的～

矩阵的图片没了……来个文字版的：

a0 1 0 0 0 0 0 0 0 0

a1 0 1 0 0 0 0 0 0 0

a2 0 0 1 0 0 0 0 0 0

a3 0 0 0 1 0 0 0 0 0

a4 0 0 0 0 1 0 0 0 0

a5 0 0 0 0 0 1 0 0 0

a6 0 0 0 0 0 0 1 0 0

a7 0 0 0 0 0 0 0 1 0

a8 0 0 0 0 0 0 0 0 1

a9 0 0 0 0 0 0 0 0 0

Solution

// HDOJ 1757
// A Simple Math Problem
// by A Code Rabbit

#include <cstdio>
#include <cstring>

const int MAXO = 12;

template <typename T>
struct Matrix {
    T e[MAXO][MAXO];
    int o;
    Matrix(int _o) { memset(e, 0, sizeof(e)); o = _o; }
    Matrix operator*(const Matrix& one) {
        Matrix res(o);
        for (int i = 0; i < o; i++)
            for (int j = 0; j < o; j++)
                for (int k = 0; k < o; k++)
                    res.e[i][j] += e[i][k] * one.e[k][j];
        return res;
    }
    Matrix operator%(int mod) {
        for (int i = 0; i < o; i++)
            for (int j = 0; j < o; j++)
                e[i][j] %= mod;
        return *this;
    }
};

template <typename T>
T QuickPower(T radix, int exp, int mod) {
    T res = radix;
    exp--;
    while (exp) {
        if (exp & 1) res = res * radix % mod;
        exp >>= 1;
        radix = radix * radix % mod;
    }
    return res;
}

int k, m;
int a[MAXO];

const int F[] = { 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 };

int main() {
    while (scanf("%d%d", &k, &m) != EOF) {
        // Input.
        for (int i = 0; i < 10; i++)
            scanf("%d", &a[i]);
        // Solve.
        if (k < 10) {
            printf("%d\n", k % m);
            continue;
        }
        Matrix<long long> one(10);
        for (int i = 0; i < one.o; i++)
            one.e[i][0] = a[i];
        for (int i = 0; i < one.o - 1; i++)
            one.e[i][i + 1] = 1;
        Matrix<long long> ans = QuickPower(one, k - 9, m);
        // Compute ans output.
        int sum = 0;
        for (int i = 0; i < 10; i++)
            sum += F[i] * ans.e[i][0];
        printf("%d\n", sum % m);
    }

    return 0;
}