hdu4767_Bell_矩阵快速幂+中国剩余定理

2013长春赛区网络赛的1009题

比赛的时候这道题英勇的挂掉了，原因是写错了一个系数，有时候粗心比脑残更可怕

本题是关于Bell数，关于Bell数的详情请见维基：http://en.wikipedia.org/wiki/Bell_number

其中有一句话是这么说的： And they satisfy "Touchard's congruence": If p is any prime bumber then

但95041567不是素数， 分解之后发现 95041567 = 31 × 37 × 41 × 43 × 47

按照上述递推式，利用矩阵快速幂可以得到 B_n mod p, (p = 31, 37, 41, 43, 47)，因为p最大47,所以矩阵快速幂O(p^3 * log(n/p))不会超时，

当然要先利用以下公式把B1-B47预处理出来：

得到5个B_n mod p (p = 31, 37, 41, 43, 47)之后，怎样得到B_n mod  Πp 呢？

利用中国剩余定理可以完美的解决上述问题

详情见代码：

#include <cstdio>

#include <cstring>



#define MOD 95041567

#define LL long long



const int maxn = 50; //必须加 const，否则编译错误

LL x[5] = {31, 37, 41, 43, 47};

LL X;



class Matrix {

public:

    LL val[maxn][maxn];

    Matrix() {

        memset(val, 0, sizeof(val));

    }



    Matrix operator*(const Matrix& c) const {

        Matrix res;

        for (int i = 0; i < X; ++i)

            for (int j = 0; j < X; ++j)

                for (int k = 0; k < X; ++k) {

                    res.val[i][j] += val[i][k] * c.val[k][j];

                    res.val[i][j] = (res.val[i][j] + X) % X; //防止矩阵元素变为负数，若不需要，去掉"+MOD"

                }

        return res;

    }



    Matrix operator*=(const Matrix& c) {

        *this = *this * c;

        return *this;

    }



    Matrix Pow(LL k) { //返回one^k

        Matrix res = Zero();

        Matrix step = One();

        while (k) {

            if (k & 1)

                res *= step;

            k >>= 1;

            step *= step;

        }

        return res;

    }



    Matrix Zero() const {

        Matrix res;

        for (int i = 0; i < X; ++i)

            res.val[i][i] = 1;

        return res;

    }



    Matrix One() const {

        Matrix res;

        for (int i = 0; i < X - 1; ++i)

            res.val[i][i] = res.val[i + 1][i] = 1;

        res.val[0][X - 1] = res.val[1][X - 1] = res.val[X - 1][X - 1] = 1;

        return res;

    }

};



void gcd(LL a, LL b, LL& d, LL& xx, LL& y) {

    if (!b) {

        d = a, xx = 1, y = 0;

    } else {

        gcd(b, a % b, d, y, xx);

        y -= xx * (a / b);

    }

}



LL china(LL n, LL* a, LL* m) {

    LL M = 1, d, xx = 0, y;

    for (int i = 0; i < n; ++i) M *= m[i];

    for (int i = 0; i < n; ++i) {

        LL w = M / m[i];

        gcd(m[i], w, d, d, y);

        xx = (xx + y * w * a[i]) % M;

    }

    return (xx + M) % M;

}



LL c[50][50], f[50], a[5];



int main() {

    int T;

    for (int i = 0; i < 50; ++i) {

        c[i][0] = c[i][i] = 1;

        for (int j = 1; j < i; ++j)

            c[i][j] = (c[i-1][j] + c[i - 1][j - 1]) % MOD;

    }

    f[0] = 1;

    f[1] = 1;

    for (int i = 2; i < 50; ++i) {

        for (int j = 0; j < i; ++j)

            f[i] = (f[i] + c[i - 1][j] * f[j]) % MOD;

    }

    scanf("%d", &T);

    while (T--) {

        LL n;

        scanf("%I64d", &n);

        if (n < 50) {

            printf("%I64d\n", f[n]);

            continue;

        }

        memset(a, 0 ,sizeof(a));

        for (int i = 0; i < 5; ++i) {

            X = x[i];

            Matrix m;

            m = m.Pow(n / X);

            for (int j = 0; j < X; ++j)

                a[i] = (a[i] + f[j] * m.val[j][n % X]) % X;

        }

        printf("%I64d\n", china(5, a, x));

    }

    return 0;

}