poj 3420 Quad Tiling

http://poj.org/problem?id=3420

　　继续矩阵快速幂！

　　题目意思是给你一个4*n的矩阵用1*2的砖块填充，问有多少种填充的方法。这题可以模仿一道老题（2*n的矩阵用1*2的砖块填充）的递推方法。不过如果是这样递推，推着推着会发现，递推式会有无穷多个项。不过从第三项开始，每项乘以的系数有规律，具体的规律还是查看代码吧！这题我找到的递推式中要分奇偶项，所以构建矩阵的时候就要划分奇偶情况来讨论，所以就构建出奇偶两种矩阵来。奇矩阵是在求奇数项的时候用的，偶矩阵同理。

此题轻松1y，代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxSize = 5;
const int initMod = 1E9 + 7;
const int Odd[5][5] = { {1, 1, 0, 0, 0},
                         {4, 0, 1, 0, 0},
                         {3, 0, 1, 0, 0},
                         {2, 0, 0, 1, 0},
                         {2, 0, 0, 0, 1} };
const int Even[5][5] = { {1, 1, 0, 0, 0},
                          {4, 0, 0, 1, 0},
                          {2, 0, 1, 0, 0},
                          {3, 0, 0, 1, 0},
                          {3, 0, 0, 0, 1}};

int curSize = maxSize;
int curMod = initMod;

struct Matrix {
    int val[maxSize][maxSize];

    Matrix (bool ONE = false) {
        for (int i = 0; i < curSize; i++) {
            for (int j = 0; j < curSize; j++) {
                val[i][j] = 0;
            }
            if (ONE) val[i][i] = 1;
        }
    }

    void print(int _l = maxSize, int _w = maxSize) {
        for (int i = 0; i < _l; i++) {
            for (int j = 0; j < _w; j++) {
                if (j) putchar(' ');
                printf("%d", val[i][j]);
            }
            puts("");
        }
    }
} odd, even, Base;

Matrix operator * (Matrix &_a, Matrix &_b) {
    Matrix ret = Matrix();

    for (int i = 0; i < curSize; i++) {
        for (int k = 0; k < curSize; k++) {
            if (_a.val[i][k]) {
                for (int j = 0; j < curSize; j++) {
                    ret.val[i][j] += _a.val[i][k] * _b.val[k][j];
                    ret.val[i][j] %= curMod;
                }
            }
        }
    }

    return ret;
}

Matrix operator ^ (Matrix &_a, int _p) {
    Matrix ret = Matrix(true);
    Matrix __a = _a;

    while (_p) {
        if (_p & 1) ret = ret * __a;
        __a = __a * __a;
        _p >>= 1;
    }
//    puts("yeah!!!");

    return ret;
}

void initMat() {
    odd = Matrix();
    even = Matrix();
    Base = Matrix();

    for (int i = 0; i < 5; i++) {
        for (int j = 0; j < 5; j++) {
            even.val[i][j] = Even[i][j];
            odd.val[i][j] = Odd[i][j];
        }
    }
    Base.val[0][0] = 5;
    Base.val[0][1] = 1;
    Base.val[0][4] = 1;
//
//    puts("odd");
//    odd.print();
//    puts("even");
//    even.print();
}

int deal(int n) {
    Matrix ans = Base;
    Matrix TWO = odd * even;

    if (n == 1) return 1 % curMod;
    if (n == 2) return 5 % curMod;
//    puts("~~TWO");
//    TWO.print();
    ans = TWO ^ ((n >> 1) - 1);
    ans = Base * ans;
    if (n & 1) ans = ans * odd;
//    puts("~ans");
//    ans.print();

    return ans.val[0][0];
}

int main() {
    int n;

      freopen("in", "r", stdin);
    initMat();
    while (~scanf("%d%d", &n, &curMod) && (n || curMod)) {
        printf("%d\n", deal(n));
    }

    return 0;
}

——written by Lyon