POJ3420 Quad Tiling

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题意不多说，直接给出公式：
f(1)=1,f(2)=5,f(3)=11,f(4)=36
n≥5时 f(n)=f(n−1)+5∗f(n−2)+f(n−3)−f(n−4)

得到递推矩阵：

⎡ ⎣ ⎢ ⎢ ⎢ ⎢ f n - 3 f n - 2 f n - 1 f n ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎢ 000 - 1 100101050014 ⎤ ⎦ ⎥ ⎥ ⎥ * ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ f n - 4 f n - 3 f n - 2 f n - 1 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥

然后用矩阵快速幂的板子飞一飞就行了。。。
要注意的是，

n≤4时，也需要对m取模，这个害我WA了半页。。。

#include <iostream>#include <cstring>using namespace std;const int MAXN = 5;const int MAXM = 5;long long mod = 0;struct Matrix {    int n, m;   //n列m行    long long a[MAXN][MAXM];    Matrix() { clear(); }    Matrix(int _n) {   //单位矩阵        clear();        n = m = _n;        for (int i = 0; i < _n; i++)            a[i][i] = 1;    }    Matrix(const Matrix &b) {        n = b.n;        m = b.m;        for (int i = 0; i < n; ++i)            for (int j = 0; j < m; ++j)                a[i][j] = b.a[i][j];    }    void clear() {        n = m = 0;        memset(a, 0, sizeof(a));    }    Matrix operator+(const Matrix &b) const {        Matrix tmp;        tmp.n = n;        tmp.m = m;        for (int i = 0; i < n; ++i)            for (int j = 0; j < m; ++j) {                tmp.a[i][j] = a[i][j] + b.a[i][j];            }        return tmp;    }    Matrix operator-(const Matrix &b) const {        Matrix tmp;        tmp.n = n;        tmp.m = m;        for (int i = 0; i < n; ++i)            for (int j = 0; j < m; ++j)                tmp.a[i][j] = a[i][j] - b.a[i][j];        return tmp;    }    Matrix operator*(const Matrix &b) const {        Matrix tmp;        tmp.clear();        tmp.n = b.n;        tmp.m = b.m;        for (int i = 0; i < n; ++i)            for (int j = 0; j < b.m; ++j) {                for (int k = 0; k < m; ++k)                    tmp.a[i][j] = (tmp.a[i][j] + ((a[i][k] % mod) * (b.a[k][j] % mod)) % mod) % mod;        }        return tmp;    }    Matrix& operator%(const long long &mod)  {        for (int i = 0; i < n; ++i)            for (int j = 0; j < m; ++j) {                a[i][j] = a[i][j]%mod;            }        return *this;    }};Matrix pow_mod(const Matrix &a, long long i, long long n) {    if (i == 0) return Matrix(4);    Matrix temp = pow_mod(a, i >> 1, n);    temp = temp * temp ;    if (i & 1) temp = temp * a ;    return temp;}//ans:(a^i)%nlong long n, m;Matrix mCoefficient, mAnsCoefficient, mF1toF4, mFn;  //系数矩阵void preWork() {    ios::sync_with_stdio(false);    cin.tie(0);    cout.tie(0);    mF1toF4.clear();    mF1toF4.n = 4;    mF1toF4.m = 1;    mF1toF4.a[0][0] = 1;    mF1toF4.a[1][0] = 5;    mF1toF4.a[2][0] = 11;    mF1toF4.a[3][0] = 36;    mCoefficient.clear();    mCoefficient.n = 4;    mCoefficient.m = 4;    mCoefficient.a[0][1] = mCoefficient.a[1][2] = mCoefficient.a[2][3]            = mCoefficient.a[3][1] = mCoefficient.a[3][3] = 1;    mCoefficient.a[3][0] = -1;    mCoefficient.a[3][2] = 5;}int main() {    preWork();    while (cin >> n >> m , n) {        if (n > 4) {            mod = m;            mAnsCoefficient = pow_mod(mCoefficient, n-4, m);            mFn = mAnsCoefficient * mF1toF4;            mFn.a[3][0]=mFn.a[3][0]%m;            while (mFn.a[3][0]<0) mFn.a[3][0]=mFn.a[3][0]+m;            cout << mFn.a[3][0] << "\n";        } else {            cout << mF1toF4.a[n - 1][0]%m << "\n";        }    }}//f(n)=f(n-1)+5*f(n-2)+f(n-3)-f(n-4)

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