DOJ 1003

已知：

\[f(1)=1\]

\[f(n)=(n-1)f(1)+(n-2)f(2)+\cdots +f(n-1)\]

求第n项的数

 

题解：

推一推，可以推出

于是

然而。。蒟蒻不会写资辞负数的高精度

所以乱搞搞

高精度搞搞就好了

 

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
typedef unsigned long long ull;
typedef vector < ull > vull;
vull :: iterator it;
const double PI = acos(-1.0);
const int N = 1000005;
struct cp {
    double r, i;
    cp (double a = 0.0, double b = 0.0) :
        r(a), i(b) {}
    inline void operator= (const int &p) {r = p, i = 0.0;}
    inline cp operator+ (const cp &o) {return cp(r + o.r, i + o.i);}
    inline cp operator- (const cp &o) {return cp(r - o.r, i - o.i);}
    inline cp operator* (const cp &o) {
        return cp(r * o.r - i * o.i, r * o.i + i * o.r);
    }
} pa[N], pb[N];

inline void rader(cp *a, int n) {
    int j = n >> 1;
    for (int i = 1; i < n - 1; ++i) {
        if (i < j) swap(a[i], a[j]);
        int k = n >> 1;
        while (j >= k) {
            j -= k; k >>= 1;
        } if (j < k) j += k;
    }
}

inline void fft(cp *a, int n, bool idft = false) {
    rader(a, n);
    double pi = idft ? -PI : PI;
    for (int h = 2; h <= n; h <<= 1) {
        cp omgn(cos(2 * pi / h), sin(2 * pi / h));
        for (int j = 0; j < n; j += h) {
            cp omg(1, 0);
            for (int k = j; k < j + h / 2; ++k) {
                cp u = a[k], v = omg * a[k + h / 2];
                a[k] = u + v, a[k + h / 2] = u - v;
                omg = omg * omgn;
            }
        }
    }
    if (idft) for (int i = 0; i < n; ++i) a[i].r /= n;
}

inline vull mul(vull &a, vull &b) {
    int la = 0, lb = 0, s = 0;
    vull res;
    for (it = a.begin(); it != a.end(); ++it)
        pa[la++] = *it;
    for (it = b.begin(); it != b.end(); ++it)
        pb[lb++] = *it;
    while ((1 << s) < la + lb) ++s; s = 1 << s;
    for (int i = la; i < s; ++i) pa[i] = 0;
    for (int i = lb; i < s; ++i) pb[i] = 0;
    fft(pa, s); fft(pb, s);
    for (int i = 0; i < s; ++i) pa[i] = pa[i] * pb[i];
    fft(pa, s, true);
    res.resize(la + lb - 1);
    for (int i = 0; i <= la + lb - 2; ++i)
        res[i] = (ull) (pa[i].r + 0.5);
    return res;
}

const int BASE = 1000, WIDTH = 3;
struct Bignum {
    vull s;
    inline Bignum operator= (ull num) {
        s.clear();
        do {
            s.push_back(num % BASE);
            num /= BASE;
        } while (num > 0);
        return *this;
    }

    inline void operator= (const string &str) {
        s.clear();
        ull x;
        int len = (str.size() - 1) / WIDTH + 1, size = str.size();
        for (int i = 0; i < len; ++i) {
            int end = size - i * WIDTH;
            int start = max(0, end - WIDTH);
            sscanf(str.substr(start, end - start).c_str(), "%llu", &x);
            s.push_back(x);
        }
        clean();
    }

    inline Bignum clean() {
        while (!s.back() and !s.empty()) {
            s.pop_back();
        }
        if (s.empty()) s.push_back(0);
        return *this;
    }

    inline void cc() {
        int size = s.size();
        for (int i = 0; i < size; ++i) {
            if (s[i] >= BASE) {
                if (i + 1 == size)
                    s.push_back(s[i] / BASE);
                else
                    s[i + 1] += s[i] / BASE;
                s[i] %= BASE;
            }
        }
    }

    inline void print() {
        printf("%llu", s.back());
        for (int i = s.size() - 2; i >= 0; --i) {
            char buf[20];
            sprintf(buf, "%03llu", s[i]);
            int len = strlen(buf);
            for (int j = 0; j < len; ++j)
                putchar(buf[j]);
        }
    }

    inline Bignum operator+ (const Bignum& o) {
        Bignum res; res.s.clear();
        int sza = s.size(), szb = o.s.size();
        ull g = 0;
        for (int i = 0;; ++i) {
            if (!g and i >= sza and i >= szb) break;
            ull x = g;
            if (i < sza) x += s[i];
            if (i < szb) x += o.s[i];
            res.s.push_back(x % BASE);
            g = x / BASE;
        }
        res.cc();
        return res;
    }

    inline Bignum& operator += (const Bignum &o) {
        *this = *this + o;
        return *this;
    }


    inline friend Bignum operator* (Bignum &a, Bignum &b) {
        Bignum res;
        res.s = mul(a.s, b.s);
        res.cc();
        return res;
    }

};

struct Mat {
    Bignum d[2][2];
    Mat(int u = 0, int i = 0, int o = 0, int p = 0) {
        d[0][0] = u; d[0][1] = i;
        d[1][0] = o; d[1][1] = p;
    }
    inline Mat operator* (Mat &o) {
        Mat res;
        for (int i = 0; i < 2; ++i)
            for (int j = 0; j < 2; ++j)
                for (int k = 0; k < 2; ++k)
                    res.d[i][j] += d[i][k] * o.d[k][j];
        return res;
    }
} A(2, 1, 1, 1), ans(1, 0, 0, 1);


int main() {
    int n; scanf("%d", &n);
    if (n < 3) {
        puts("1"); return 0;
    }
    if (n == 3) {
        puts("3"); return 0;
    }
    --n;
    while (n) {
        if (n & 1) ans = A * ans;
        A = A * A; n >>= 1;
    }
    ans.d[0][1].print();
    return 0;
}