HDU 1588 Gauss Fibonacci 矩阵

首先fib数列可以很随意的推出来矩阵解法，然后这里就是要处理一个关于矩阵的等比数列求和的问题，这里有一个logn的解法,类似与这样

A^0+A^1+A^2+A^3 = A^0 + A^1 + A^2 * (A^0 + A^1) 处理就好了。

#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 = 6;
LL k, b, n, 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) {
    int n = a.n, m = b.m;
    Matrix ret(n, m);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= 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 operator + (Matrix a, Matrix b) {
    for(int i = 1; i <= a.n; i++) {
        for(int j = 1; j <= a.m; j++) {
            a.data[i][j] += b.data[i][j];
            a.data[i][j] %= mod;
        }
    }
    return a;
}

Matrix pow(Matrix mat, LL k) {
    if(k == 0) {
        Matrix ret(mat.n, mat.m);
        for(int i = 1; i <= mat.n; i++) ret.data[i][i] = i;
        return ret;
    }
    if(k == 1) return mat;
    Matrix ret = pow(mat * mat, k / 2);
    if(k & 1) ret = ret * mat;
    return ret;
}

Matrix calc(Matrix mat, LL p) {
    if(p == 0) return pow(mat, b);
    if(p == 1) return pow(mat, b) + pow(mat, k + b);
    int midval = (p + 1) % 2 == 0 ? (p / 2) : (p / 2) - 1;
    Matrix ret = calc(mat, midval);
    ret = ret + pow(mat, (midval + 1) * k) * ret;
    if((p + 1) & 1) ret = ret + pow(mat, p * k + b);
    return ret;
}

int main() {
    while(cin >> k >> b >> n >> mod) {
        Matrix A(2, 2);
        A.data[1][1] = A.data[1][2] = A.data[2][1] = 1;
        A.data[2][2] = 0;
        Matrix ans = calc(A, n - 1);
        cout << ans.data[2][1] << endl;
    }
    return 0;
}