HDOJ 1588 - Gauss Fibonacci

Matrix Multiplication (& Quick Power) 

Description

g(x) = k * x + b。

f(x) 为Fibonacci数列。

求f(g(x))，从x = 1到n的数字之和，并对m取模。

Type

Matrix Multiplication

Quick Power

Analysis

我们知道f(x)中，两个元素之间的关系是

       

因为g(x) – b，为一个等比数列，所以，他们之间也有一个类似Fibonacci的关系，并且可以用矩阵来表示

而为了求和，我们可以添加一项s(x)，用来表示前x项的和，最后得到矩阵

( s(x-1),f(g(x)),f(g(x)-1) ) = ( s(x – 2), f(g(x – 1), f(g(x – 1) – 1)) ) ×

 

剩下的工作就是矩阵乘法和快速幂了。

Solution

// HDOJ 1588// Gauss Fibonacci// by A Code Rabbit#include <cstdio>#include <cstring>const int MAXO = 5;template <typename T>struct Matrix {    T e[MAXO][MAXO];    int o;    Matrix(int order) { memset(e, 0, sizeof(e)); o = order; }    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, b, n, m;int Fibonacci(int x);int main() {    while (scanf("%d%d%d%d", &k, &b, &n, &m) != EOF) {        // Initialize in the first time.        Matrix<long long> mat_one1(2);        mat_one1.e[0][0] = 1;        mat_one1.e[0][1] = 1;        mat_one1.e[1][0] = 1;        // Quick power for computing the matrix of the relation of two nest        // pair of numbers in g(x).        Matrix<long long> mat_ans1 = QuickPower(mat_one1, k, m);        // Initialize in the second time.        Matrix<long long> mat_one2(3);        mat_one2.e[0][0] = 1;        mat_one2.e[1][0] = 1;        for (int i = 1; i < 3; ++i)            for (int j = 1; j < 3; ++j)                mat_one2.e[i][j] = mat_ans1.e[i - 1][j - 1];        // Quick power for computing the sum of g(x).        Matrix<long long> mat_ans2 = QuickPower(mat_one2, n - 1, m);        // Output.        int original_solution[] = {            Fibonacci(b),            Fibonacci(k + b),            Fibonacci(k + b - 1),        };        long long sum = 0;        for (int i = 0; i < 3; i++)            sum += original_solution[i] * mat_ans2.e[i][0];        printf("%lld\n", sum % m);    }    return 0;}int Fibonacci(int x) {    if (!x) return 0;    Matrix<long long> mat_one(2);    // Initialize mat_one.    mat_one.e[0][0] = 1;    mat_one.e[0][1] = 1;    mat_one.e[1][0] = 1;    // Quick power.    Matrix<long long> mat_ans = QuickPower(mat_one, x, m);    // Compete and return the result.    return mat_ans.e[1][0];}