作者： officeyutong

CFGYM103415A CCPC2021广州A Math Ball

2021年11月26日 by officeyutong·一条评论

题意: 有$n$种球，每种有无限个，同时第$i$种球有一个代价$c_i$，你要拿不超过$w$个球。如果最后第$i$种球你拿了$k_i$个，那么你会获得$\prod_{1\leq 阅读全文..

跑的比较快的多项式板子

2021年11月13日 by officeyutong·0评论

#include <cstring>
#include <iostream>
#include <random>
using int_t = long long int;
using std::cin;
using std::cout;
using std::endl;

const int_t LARGE = 2e6;

const int_t mod = 998244353;
const int_t g = 3;
int_t power(int_t b, int_t i) {
    int_t r = 1;
    if (i < 0)
        i = (i % (mod - 1) + mod - 1) % (mod - 1);
    while (i) {
        if (i & 1)
            r = r * b % mod;
        b = b * b % mod;
        i >>= 1;
    }
    return r;
}

void makeflip(int_t* arr, int_t size2) {
    int_t len = (1 << size2);
    arr[0] = 0;
    for (int_t i = 1; i < len; i++) {
        arr[i] = (arr[i >> 1] >> 1) | ((i & 1) << (size2 - 1));
    }
}

int_t upper2n(int_t x) {
    int_t r = 0;
    while ((1 << r) < x)
        r++;
    return r;
}
template <int_t arg = 1>
void transform(int_t* A, int_t size2, int_t* flip) {
    int_t len = (1 << size2);
    for (int_t i = 0; i < len; i++) {
        int_t r = flip[i];
        if (r > i)
            std::swap(A[i], A[r]);
    }
    for (int_t i = 2; i <= len; i *= 2) {
        int_t mr = power(g, arg * (mod - 1) / i);
        for (int_t j = 0; j < len; j += i) {
            int_t curr = 1;
            for (int_t k = 0; k < i / 2; k++) {
                int_t u = A[j + k], v = curr * A[j + k + i / 2] % mod;
                A[j + k] = (u + v) % mod;
                A[j + k + i / 2] = (u - v + mod) % mod;
                curr = curr * mr % mod;
            }
        }
    }
}
void poly_inv(const int_t* A, int_t n, int_t* result) {
    /*
    这里的n是模x^n
    计算B(x)*A(x) = 1 mod x^n, 其中A(x)已知
    假设已知A(x)*B(x) = 1 mod x^{ceil(n/2)}
    假设C(x)*A(x) = 1 mod x^n
    (A(x)B(x)-1)^2 = A^2(x)B^2(x)-2A(x)B(x)+1= 0
    A(x)B^2(x)-2B(x)+C(x) = 0
    C(x) = 2B(x)-A(x)B^2(x)
    */
    // #ifdef DEBUG
    //     cout << "at " << n << endl;
    // #endif
    if (n == 1) {
        result[0] = power(A[0], -1);
        return;
    }
    int_t next = n / 2 + n % 2;
    poly_inv(A, next, result);
    //次数不要选错了，应该用n次的A和B去卷
    int_t size2 = upper2n(n + 2 * next + 1);
    static int_t X[LARGE];
    static int_t Y[LARGE];
    int_t len = (1 << size2);
    // 写错设置范围了！
    memset(X + n, 0, sizeof(X[0]) * (len - n));
    memset(Y + next, 0, sizeof(Y[0]) * (len - next));
    memcpy(X, A, sizeof(A[0]) * n);
    memcpy(Y, result, sizeof(result[0]) * next);
    static int_t fliparr[LARGE];
    makeflip(fliparr, size2);
    transform<1>(X, size2, fliparr);
    transform<1>(Y, size2, fliparr);

    for (int_t i = 0; i < len; i++) {
        X[i] = (2 * Y[i] - X[i] * Y[i] % mod * Y[i] % mod + mod) % mod;
    }
    transform<-1>(X, size2, fliparr);
    const int_t inv = power(len, -1);
    for (int_t i = 0; i < n; i++)
        result[i] = X[i] * inv % mod;
#ifdef DEBUG
    cout << "poly inv at " << n << endl;
    cout << "result = ";
    for (int_t i = 0; i < next; i++)
        cout << result[i] << " ";
    cout << endl;

#endif
}
int_t poly_divat(const int_t* F, const int_t* G, int_t n, int_t k) {
    /*
        n次多项式F和G
        计算F(x)/G(x)的k次项前系数
        考虑F(x)*G(-x)/G(x)*G(-x)，分母只有偶数次项，写为C(x^2)；分子写成xA(x^2)+B(x^2)，如果k是奇数，那么递归(A,C,n,k/2)，如果k是偶数，那么递归(B,C,n,k/2)
        到k<=n时直接计算
    */
    int_t size2 = upper2n(2 * n + 1);
    int_t len = 1 << size2;
    static int_t fliparr[LARGE];
    makeflip(fliparr, size2);
    static int_t T1[LARGE], T2[LARGE], T3[LARGE];
    for (int_t i = 0; i < len; i++) {
        if (i <= n)
            T1[i] = F[i], T2[i] = G[i];
        else
            T1[i] = T2[i] = T3[i] = 0;
    }
    const int_t inv = power(len, -1);
    while (k >= n) {
#ifdef DEBUG
        cout << "curr at k = " << k << endl;
#endif
        for (int_t i = 0; i < len; i++) {
            if (i <= n) {
                T3[i] = T2[i] * (i % 2 ? (mod - 1) : 1);
            } else
                T3[i] = 0;
        }
        transform(T1, size2, fliparr);
        transform(T2, size2, fliparr);
        transform(T3, size2, fliparr);
        for (int_t i = 0; i < len; i++) {
            T1[i] = T1[i] * T3[i] % mod;
            T2[i] = T2[i] * T3[i] % mod;
        }
        transform<-1>(T1, size2, fliparr);
        transform<-1>(T2, size2, fliparr);
#ifdef DEBUG
        cout << "prod T1 = ";
        for (int_t i = 0; i < len; i++)
            cout << T1[i] * inv % mod << " ";
        cout << endl;
        cout << "prod T2 = ";
        for (int_t i = 0; i < len; i++)
            cout << T2[i] * inv % mod << " ";
        cout << endl;

#endif
        for (int_t i = 0; i < len; i++) {
            if (i * 2 < len) {
                T2[i] = T2[i * 2] * inv % mod;
            } else
                T2[i] = 0;
        }
        int_t b = k % 2;
        for (int_t i = 0; i < len; i++) {
            if (i % 2 == b) {
                T1[i / 2] = T1[i] * inv % mod;
#ifdef DEBUG
                cout << "at " << i << " assgin " << T1[i] * inv << " to "
                     << i / 2 << endl;
#endif
            }
#ifdef DEBUG
            cout << "assign 0 to " << i << endl;
#endif
            if (i > 0)
                T1[i] = 0;
        }
#ifdef DEBUG
        cout << "finished T1 = ";
        for (int_t i = 0; i < len; i++)
            cout << T1[i] % mod << " ";
        cout << endl;
        cout << "finished T2 = ";
        for (int_t i = 0; i < len; i++)
            cout << T2[i] % mod << " ";
        cout << endl;

#endif
        k >>= 1;
    }

    poly_inv(T2, k + 1, T3);
    // #ifdef DEBUG

    //     cout << "finished k = " << k << endl;
    //     cout << "T1 = ";
    //     for (int_t i = 0; i < len; i++)
    //         cout << T1[i] % mod << " ";
    //     cout << endl;
    //     cout << "T2 = ";
    //     for (int_t i = 0; i < len; i++)
    //         cout << T2[i] % mod << " ";
    //     cout << endl;
    //     cout << "T2 inv = ";
    //     for (int_t i = 0; i < len; i++)
    //         cout << T3[i] % mod << " ";
    //     cout << endl;

    // #endif
    int_t result = 0;
    //计算结果的k次项
    for (int_t i = 0; i <= k; i++) {
        result = (result + T1[i] * T3[k - i] % mod) % mod;
    }
    return result;
}
void poly_mul(const int_t* A, int_t n, const int_t* B, int_t m, int_t* C) {
    /*
       计算n次多项式A与m次多项式B的乘积
    */
    int_t size2 = upper2n(n + m + 1);
    int_t len = 1 << size2;
    static int_t T1[LARGE], T2[LARGE];
    for (int_t i = 0; i < len; i++) {
        if (i <= n)
            T1[i] = A[i];
        else
            T1[i] = 0;
        if (i <= m)
            T2[i] = B[i];
        else
            T2[i] = 0;
    }
    static int_t fliparr[LARGE];
    makeflip(fliparr, size2);
    transform(T1, size2, fliparr);
    transform(T2, size2, fliparr);
    for (int_t i = 0; i < len; i++)
        T1[i] = T1[i] * T2[i] % mod;
    transform<-1>(T1, size2, fliparr);
    int_t inv = power(len, -1);
    for (int_t i = 0; i <= n + m; i++)
        C[i] = T1[i] * inv % mod;
}
int_t poly_linear_rec(const int_t* A0, const int_t* F0, int_t n, int_t k) {
    /*
        计算线性递推
        F[1],F[2]...F[k] 递推系数
        A[0],A[1]...A[k-1] 首项

        A[m]=A[m-1]*F[1]+A[m-2]*F[2]+...+A[m-k]*F[k]
    */

    static int_t T1[LARGE], T2[LARGE];
    static int_t Ax[LARGE], Fx[LARGE];
    Fx[0] = 0;
    Ax[k] = 0;
    for (int i = 1; i <= k; i++) {
        Fx[i] = F0[i];
        Ax[i - 1] = A0[i - 1];
    }
    poly_mul(Ax, k, Fx, k, T1);
    T1[0] = Ax[0];
    for (int i = 1; i <= k - 1; i++) {
        T1[i] = (Ax[i] - T1[i] + mod) % mod;
    }
    for (int i = k; i <= 2 * k; i++)
        T1[i] = 0;
    T2[0] = 1;
    for (int i = 1; i <= k; i++)
        T2[i] = (mod - Fx[i]) % mod;

    // #ifdef DEBUG
    //     cout << "T1 = ";
    //     for (int_t i = 0; i <= k; i++) {
    //         cout << T1[i] << " ";
    //     }
    //     cout << endl;
    //     cout << "T2 = ";
    //     for (int_t i = 0; i <= k; i++) {
    //         cout << T2[i] << " ";
    //     }
    //     cout << endl;

    // #endif
    return poly_divat(T1, T2, k, n);
}

void poly_log(const int_t* A, int_t n, int_t* out) {
    /*
        计算log(A(x)), A(x)为n次多项式
        DlogF(x) =DF(x)/F(x)
    */
    static int_t Ad[LARGE];
    static int_t Finv[LARGE];
    static int_t R[LARGE];
    const int_t m = n - 1;
    for (int_t i = 0; i <= m; i++) {
        Ad[i] = A[i + 1] * (i + 1) % mod;
    }
    Ad[n] = 0;
    poly_inv(A, n + 1, Finv);
    poly_mul(Ad, m, Finv, n, R);
    for (int_t i = 1; i <= n; i++) {
        out[i] = R[i - 1] * power(i, -1) % mod;
    }
}
void poly_exp(const int_t* A, int_t n, int_t* out) {
    /*
        计算exp(A(x)), A(x)为n次多项式
        H(x)=G(x)(1-logG(x)+A(x))
        H(x)为当次递推项，G(x)为上一次递推项
    */
    if (n == 1) {
        out[0] = 1;
        return;
    }
    int_t r = n / 2 + n % 2;
    poly_exp(A, r, out);
    static int_t G[LARGE];
    static int_t G2[LARGE];
    static int_t logG[LARGE];
    static int_t R[LARGE];
    for (int_t i = 0; i < r; i++)
        G[i] = out[i];
    for (int_t i = r; i < n; i++)
        G[i] = 0;
    poly_log(G, n - 1, logG);
    // for (int_t i = r; i < n; i++)
    //     logG[i] = 0;
    for (int_t i = 0; i < n; i++) {
        G2[i] = (mod - logG[i] + A[i]) % mod;
    }
    G2[0] = (G2[0] + 1) % mod;
    poly_mul(G, n - 1, G2, n - 1, R);
    for (int_t i = 0; i < n; i++)
        out[i] = R[i];
#ifdef DEBUG
    cout << "at " << n << endl;
    cout << "A = ";
    for (int_t i = 0; i < n; i++)
        cout << A[i] << " ";
    cout << endl;
    cout << "G = ";
    for (int_t i = 0; i < n; i++)
        cout << G[i] << " ";
    cout << endl;
    cout << "logG = ";
    for (int_t i = 0; i < n; i++)
        cout << logG[i] << " ";
    cout << endl;
    cout << "G2 = ";
    for (int_t i = 0; i < n; i++)
        cout << G2[i] << " ";
    cout << endl;
    cout << "R = ";
    for (int_t i = 0; i < n; i++)
        cout << R[i] << " ";
    cout << endl;
    cout << endl;
#endif
}
int main() {
    std::ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int_t n;
    cin >> n;
    static int_t A[LARGE], B[LARGE];
    static int_t C[LARGE];
    for (int_t i = 0; i < n; i++)
        cin >> A[i];
    poly_exp(A, n, B);
    for (int_t i = 0; i < n; i++)
        cout << B[i] << " ";

    return 0;
}

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#include <cstring>

#include <iostream>

#include <random>

using int_t = long long int;

using std::cin;

using std::cout;

using std::endl;

const int_t LARGE = 2e6;

const int_t mod = 998244353;

const int_t g = 3;

int_t power(int_t b, int_t i) {

int_t r = 1;

if (i < 0)

i = (i % (mod - 1) + mod - 1) % (mod - 1);

while (i) {

if (i & 1)

r = r * b % mod;

b = b * b % mod;

i >>= 1;

}

return r;

}

void makeflip(int_t* arr, int_t size2) {

int_t len = (1 << size2);

arr[0] = 0;

for (int_t i = 1; i < len; i++) {

arr[i] = (arr[i >> 1] >> 1) | ((i & 1) << (size2 - 1));

}

int_t upper2n(int_t x) {

int_t r = 0;

while ((1 << r) < x)

r++;

return r;

}

template <int_t arg = 1>

void transform(int_t* A, int_t size2, int_t* flip) {

int_t len = (1 << size2);

for (int_t i = 0; i < len; i++) {

int_t r = flip[i];

if (r > i)

std::swap(A[i], A[r]);

}

for (int_t i = 2; i <= len; i *= 2) {

int_t mr = power(g, arg * (mod - 1) / i);

for (int_t j = 0; j < len; j += i) {

int_t curr = 1;

for (int_t k = 0; k < i / 2; k++) {

int_t u = A[j + k], v = curr * A[j + k + i / 2] % mod;

A[j + k] = (u + v) % mod;

A[j + k + i / 2] = (u - v + mod) % mod;

curr = curr * mr % mod;

}

void poly_inv(const int_t* A, int_t n, int_t* result) {

这里的n是模x^n

计算B(x)*A(x) = 1 mod x^n, 其中A(x)已知

假设已知A(x)*B(x) = 1 mod x^{ceil(n/2)}

假设C(x)*A(x) = 1 mod x^n

(A(x)B(x)-1)^2 = A^2(x)B^2(x)-2A(x)B(x)+1= 0

A(x)B^2(x)-2B(x)+C(x) = 0

C(x) = 2B(x)-A(x)B^2(x)

// #ifdef DEBUG

// cout << "at " << n << endl;

// #endif

if (n == 1) {

result[0] = power(A[0], -1);

return;

}

int_t next = n / 2 + n % 2;

poly_inv(A, next, result);

//次数不要选错了，应该用n次的A和B去卷

int_t size2 = upper2n(n + 2 * next + 1);

static int_t X[LARGE];

static int_t Y[LARGE];

int_t len = (1 << size2);

// 写错设置范围了！

memset(X + n, 0, sizeof(X[0]) * (len - n));

memset(Y + next, 0, sizeof(Y[0]) * (len - next));

memcpy(X, A, sizeof(A[0]) * n);

memcpy(Y, result, sizeof(result[0]) * next);

static int_t fliparr[LARGE];

makeflip(fliparr, size2);

transform<1>(X, size2, fliparr);

transform<1>(Y, size2, fliparr);

for (int_t i = 0; i < len; i++) {

X[i] = (2 * Y[i] - X[i] * Y[i] % mod * Y[i] % mod + mod) % mod;

}

transform<-1>(X, size2, fliparr);

const int_t inv = power(len, -1);

for (int_t i = 0; i < n; i++)

result[i] = X[i] * inv % mod;

#ifdef DEBUG

cout << "poly inv at " << n << endl;

cout << "result = ";

for (int_t i = 0; i < next; i++)

cout << result[i] << " ";

cout << endl;

#endif

}

int_t poly_divat(const int_t* F, const int_t* G, int_t n, int_t k) {

n次多项式F和G

计算F(x)/G(x)的k次项前系数

考虑F(x)*G(-x)/G(x)*G(-x)，分母只有偶数次项，写为C(x^2)；分子写成xA(x^2)+B(x^2)，如果k是奇数，那么递归(A,C,n,k/2)，如果k是偶数，那么递归(B,C,n,k/2)

到k<=n时直接计算

int_t size2 = upper2n(2 * n + 1);

int_t len = 1 << size2;

static int_t fliparr[LARGE];

makeflip(fliparr, size2);

static int_t T1[LARGE], T2[LARGE], T3[LARGE];

for (int_t i = 0; i < len; i++) {

if (i <= n)

T1[i] = F[i], T2[i] = G[i];

else

T1[i] = T2[i] = T3[i] = 0;

}

const int_t inv = power(len, -1);

while (k >= n) {

#ifdef DEBUG

cout << "curr at k = " << k << endl;

#endif

for (int_t i = 0; i < len; i++) {

if (i <= n) {

T3[i] = T2[i] * (i % 2 ? (mod - 1) : 1);

} else

T3[i] = 0;

}

transform(T1, size2, fliparr);

transform(T2, size2, fliparr);

transform(T3, size2, fliparr);

for (int_t i = 0; i < len; i++) {

T1[i] = T1[i] * T3[i] % mod;

T2[i] = T2[i] * T3[i] % mod;

}

transform<-1>(T1, size2, fliparr);

transform<-1>(T2, size2, fliparr);

#ifdef DEBUG

cout << "prod T1 = ";

for (int_t i = 0; i < len; i++)

cout << T1[i] * inv % mod << " ";

cout << endl;

cout << "prod T2 = ";

for (int_t i = 0; i < len; i++)

cout << T2[i] * inv % mod << " ";

cout << endl;

#endif

for (int_t i = 0; i < len; i++) {

if (i * 2 < len) {

T2[i] = T2[i * 2] * inv % mod;

} else

T2[i] = 0;

}

int_t b = k % 2;

for (int_t i = 0; i < len; i++) {

if (i % 2 == b) {

T1[i / 2] = T1[i] * inv % mod;

#ifdef DEBUG

cout << "at " << i << " assgin " << T1[i] * inv << " to "

<< i / 2 << endl;

#endif

}

#ifdef DEBUG

cout << "assign 0 to " << i << endl;

#endif

if (i > 0)

T1[i] = 0;

}

#ifdef DEBUG

cout << "finished T1 = ";

for (int_t i = 0; i < len; i++)

cout << T1[i] % mod << " ";

cout << endl;

cout << "finished T2 = ";

for (int_t i = 0; i < len; i++)

cout << T2[i] % mod << " ";

cout << endl;

#endif

k >>= 1;

}

poly_inv(T2, k + 1, T3);

// #ifdef DEBUG

// cout << "finished k = " << k << endl;

// cout << "T1 = ";

// for (int_t i = 0; i < len; i++)

// cout << T1[i] % mod << " ";

// cout << endl;

// cout << "T2 = ";

// for (int_t i = 0; i < len; i++)

// cout << T2[i] % mod << " ";

// cout << endl;

// cout << "T2 inv = ";

// for (int_t i = 0; i < len; i++)

// cout << T3[i] % mod << " ";

// cout << endl;

// #endif

int_t result = 0;

//计算结果的k次项

for (int_t i = 0; i <= k; i++) {

result = (result + T1[i] * T3[k - i] % mod) % mod;

}

return result;

}

void poly_mul(const int_t* A, int_t n, const int_t* B, int_t m, int_t* C) {

计算n次多项式A与m次多项式B的乘积

int_t size2 = upper2n(n + m + 1);

int_t len = 1 << size2;

static int_t T1[LARGE], T2[LARGE];

for (int_t i = 0; i < len; i++) {

if (i <= n)

T1[i] = A[i];

else

T1[i] = 0;

if (i <= m)

T2[i] = B[i];

else

T2[i] = 0;

}

static int_t fliparr[LARGE];

makeflip(fliparr, size2);

transform(T1, size2, fliparr);

transform(T2, size2, fliparr);

for (int_t i = 0; i < len; i++)

T1[i] = T1[i] * T2[i] % mod;

transform<-1>(T1, size2, fliparr);

int_t inv = power(len, -1);

for (int_t i = 0; i <= n + m; i++)

C[i] = T1[i] * inv % mod;

}

int_t poly_linear_rec(const int_t* A0, const int_t* F0, int_t n, int_t k) {

计算线性递推

F[1],F[2]...F[k] 递推系数

A[0],A[1]...A[k-1] 首项

A[m]=A[m-1]*F[1]+A[m-2]*F[2]+...+A[m-k]*F[k]

static int_t T1[LARGE], T2[LARGE];

static int_t Ax[LARGE], Fx[LARGE];

Fx[0] = 0;

Ax[k] = 0;

for (int i = 1; i <= k; i++) {

Fx[i] = F0[i];

Ax[i - 1] = A0[i - 1];

}

poly_mul(Ax, k, Fx, k, T1);

T1[0] = Ax[0];

for (int i = 1; i <= k - 1; i++) {

T1[i] = (Ax[i] - T1[i] + mod) % mod;

}

for (int i = k; i <= 2 * k; i++)

T1[i] = 0;

T2[0] = 1;

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

T2[i] = (mod - Fx[i]) % mod;

// #ifdef DEBUG

// cout << "T1 = ";

// for (int_t i = 0; i <= k; i++) {

// cout << T1[i] << " ";

// }

// cout << endl;

// cout << "T2 = ";

// for (int_t i = 0; i <= k; i++) {

// cout << T2[i] << " ";

// }

// cout << endl;

// #endif

return poly_divat(T1, T2, k, n);

}

void poly_log(const int_t* A, int_t n, int_t* out) {

计算log(A(x)), A(x)为n次多项式

DlogF(x) =DF(x)/F(x)

static int_t Ad[LARGE];

static int_t Finv[LARGE];

static int_t R[LARGE];

const int_t m = n - 1;

for (int_t i = 0; i <= m; i++) {

Ad[i] = A[i + 1] * (i + 1) % mod;

}

Ad[n] = 0;

poly_inv(A, n + 1, Finv);

poly_mul(Ad, m, Finv, n, R);

for (int_t i = 1; i <= n; i++) {

out[i] = R[i - 1] * power(i, -1) % mod;

}

void poly_exp(const int_t* A, int_t n, int_t* out) {

计算exp(A(x)), A(x)为n次多项式

H(x)=G(x)(1-logG(x)+A(x))

H(x)为当次递推项，G(x)为上一次递推项

if (n == 1) {

out[0] = 1;

return;

}

int_t r = n / 2 + n % 2;

poly_exp(A, r, out);

static int_t G[LARGE];

static int_t G2[LARGE];

static int_t logG[LARGE];

static int_t R[LARGE];

for (int_t i = 0; i < r; i++)

G[i] = out[i];

for (int_t i = r; i < n; i++)

G[i] = 0;

poly_log(G, n - 1, logG);

// for (int_t i = r; i < n; i++)

// logG[i] = 0;

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

G2[i] = (mod - logG[i] + A[i]) % mod;

}

G2[0] = (G2[0] + 1) % mod;

poly_mul(G, n - 1, G2, n - 1, R);

for (int_t i = 0; i < n; i++)

out[i] = R[i];

#ifdef DEBUG

cout << "at " << n << endl;

cout << "A = ";

for (int_t i = 0; i < n; i++)

cout << A[i] << " ";

cout << endl;

cout << "G = ";

for (int_t i = 0; i < n; i++)

cout << G[i] << " ";

cout << endl;

cout << "logG = ";

for (int_t i = 0; i < n; i++)

cout << logG[i] << " ";

cout << endl;

cout << "G2 = ";

for (int_t i = 0; i < n; i++)

cout << G2[i] << " ";

cout << endl;

cout << "R = ";

for (int_t i = 0; i < n; i++)

cout << R[i] << " ";

cout << endl;

#endif

}

int main() {

std::ios::sync_with_stdio(false);

cin.tie(0), cout.tie(0);

int_t n;

cin >> n;

static int_t A[LARGE], B[LARGE];

static int_t C[LARGE];

for (int_t i = 0; i < n; i++)

cin >> A[i];

poly_exp(A, n, B);

for (int_t i = 0; i < n; i++)

cout << B[i] << " ";

return 0;

}

常系数线性递推的新做法 – 计算[x^k]P(x)/Q(x)

2021年8月12日 by officeyutong·0评论

$$ a_n=\sum_{1\le i\le k}{f_ia_{n-i}} \\ \left\{ a_0,a_1....a_{k-1} \right\} \text{已知} \\ \\ \text{设}F\left( 阅读全文..

Nginx路由匹配模拟器

2021年5月26日 by officeyutong·0评论

https://nginx.viraptor.info/

真是我的救星，调路由调吐了

CF1521C Nastia and a Hidden Permutation

2021年5月10日 by officeyutong·0评论

真是个沙比提，比赛的时候也想出来了，但是因为太迷糊了没写完

阅读全文..

CFGYM102394L CCPC2019哈尔滨 L题 LRU Algorithm

2021年4月22日 by officeyutong·0评论

阅读全文..

CFGYM102394E CCPC2019哈尔滨 E题 Exchanging Gifts

2021年4月21日 by officeyutong·0评论

假设我们能维护出最终序列的长度L和最终序列出现最多的数的个数cnt，假设$cnt\leq\frac L 2$ ，那么答案是L,这时候我们把序列升序和降序对起来就构造出n的结果。假设$cnt>\frac L 2$,那么答案是$2(L-cnt)$，我们仍然把序列升序和降序对应起来，答案是$L-(cnt-(L-cnt))=2*(L-cnt)$ 比如 1 2 3 3 3 3 3 3 3 3 3 3 2 1 中间有3个3是重了的，那么重的部分有cnt-(L-cnt)个，其中L-cnt表示的是1 2的长度，所以总答案是L-(cnt-(L-cnt)) 现在的问题在于如何维护出这个众数，并且判定$cnt\leq \frac L 2$是否成立。考虑一种线性时间求求序列众数(出现次数大于一半)的方法：令f(i)表示我们从头开始扫到第i个元素的时候(用$a_i$表示)，序列中出现次数最多的数与其他的数出现次数之和的差值。同时我们需要维护这个数是多少(用x来表示)。每次扫到$a_i$的时候: - 如果$a_i=x$，那么f(i)=f(i-1)+1,x不变 - 如果$a_i\neq x$,那么f(i)=f(i-1)-1，然后如果$f(i)<0$,那么x变为$a_i$，同时$f(i)$取反。对于这个题，如果我们要使用这种求众数的方法，核心在于如何考虑操作2（合并两个序列时）如何处理。我们维护每个序列的f和x，合并两个序列的时候： - 如果他们的f相同，那么新序列的f不变，x相加。 - 如果他们的f不同，那么考虑下两个序列的哪一个的x比较大。如果他们相同，那么新序列的f就写为0(这时候可能存在两个众数)，x从原来的x里随便选一个。如果他们不同，那x选择为f较大的那一个，同时f设置为较大值减掉较小值。所以对于这个题，我们先用这种求众数的方法算出来最终序列的众数。但此时求出来的众数，仅仅是在保证存在一个出现次数超过序列长度一半时候的众数，如果不存在这样子的众数，也会得出来一个结果，但是并不具有意义。所以我们再跑一遍递推，求出来我们上一步求的众数的出现次数，然后我们就可以照着前文所述来算答案了。 *此外，这个题卡常* 我跑了半天性能分析后大概知道了几个卡常的点: 1. 输入量非常巨大，请考虑一次性读入输入数据后自行用缓冲区处理输入。 2. std::vector的构造函数非常慢(性能分析显示,$10^6$次调用大约花了80ms)，所以不要使用vector来存储变长的序列，考虑自行分配-回收内存。 3. State结构体的构造函数占了大概40ms的时间，考虑强制内联。另外，读入函数千万不要写错了！不要写成chr>='9'！

#pragma GCC optimize("O2")
#include <assert.h>
#include <inttypes.h>
#include <algorithm>
#include <iostream>
#include <vector>


using int_t = int;
using std::cin;
using std::cout;
using std::endl;
const int_t LARGE = 1e6;
char inputbuf[(int(2e8) / 1024) * 1024];
char* head = inputbuf;
inline char nextchar() {
    return *(head++);
}
void initinput() {
    fread(inputbuf, 1024, sizeof(inputbuf) / 1024, stdin);
}
struct State {
    int64_t len;
    int_t mostval;
    int64_t mostcount;
    inline State(int64_t len = 0, int_t mostval = 0, int64_t mostcount = 0)
        : len(len), mostval(mostval), mostcount(mostcount) {}
    State operator+(const State& rhs) const {
        State result(len + rhs.len, 0, 0);
        if (mostval == rhs.mostval)
            result.mostval = mostval,
            result.mostcount = mostcount + rhs.mostcount;
        else {
            if (mostcount > rhs.mostcount) {
                result.mostcount = mostcount - rhs.mostcount;
                result.mostval = mostval;
            } else {
                result.mostcount = rhs.mostcount - mostcount;
                result.mostval = rhs.mostval;
            }
        }
        return result;
    }
};
std::ostream& operator<<(std::ostream& os, const State& state) {
    os << "State{len=" << state.len << ",mostval=" << state.mostval
       << ",mostcount=" << state.mostcount << "}";
    return os;
}
struct Opt {
    int_t type;
    int_t* data;
    int_t datalen;
    int_t x1, x2;
    int64_t mostcount = 0;
} opts[LARGE + 1];
State dp[LARGE + 1];
int_t n;
template <class T>
void read(T& x) {
    x = 0;
    char chr = nextchar();
    while (chr < '0' || chr > '9')
        chr = nextchar();
    while (chr >= '0' && chr <= '9') {
        x = x * 10 + chr - '0';
        chr = nextchar();
    }
    assert(x >= 0);
}
template <class T>
void write(T x) {
    assert(x >= 0);
    if (x > 9)
        write(x / 10);
    putchar('0' + x % 10);
}
int main() {
    // freopen("input.txt", "r", stdin);
    initinput();
    int_t T;
    read(T);
    while (T--) {
        read(n);
        for (int_t i = 1; i <= n; i++) {
            auto& ref = opts[i];
            if (ref.data) {
                delete[] ref.data;
                ref.data = nullptr;
            }
            dp[i] = State();
            read(ref.type);
            if (ref.type == 1) {
                int_t k;
                read(k);
                int_t sum = 0, val = 0;
                ref.data = new int_t[k + 1];
                ref.datalen = k;
                for (int_t i = 1; i <= k; i++) {
                    int_t x;
                    read(x);
                    ref.data[i] = x;
                    if (x == val)
                        sum++;
                    else
                        sum--;
                    if (sum < 0) {
                        sum *= -1, val = x;
                    }
                }
                // ref.seq.shrink_to_fit();
                dp[i] = State(k, val, sum);
            } else {
                read(ref.x1);
                read(ref.x2);
            }
        }
        for (int_t i = 1; i <= n; i++) {
            const auto& ref1 = opts[i];
            if (ref1.type == 2) {
                dp[i] = dp[ref1.x1] + dp[ref1.x2];
            }
        }
#ifdef DEBUG
        for (int_t i = 1; i <= n; i++) {
            cout << "dp " << i << " = " << dp[i] << endl;
        }
#endif
        int_t mostval = dp[n].mostval;
        for (int_t i = 1; i <= n; i++) {
            auto& ref = opts[i];
            if (ref.type == 1)
                ref.mostcount = std::count(ref.data + 1,
                                           ref.data + 1 + ref.datalen, mostval);
            else
                ref.mostcount = opts[ref.x1].mostcount + opts[ref.x2].mostcount;
        }
        int64_t mostcount = opts[n].mostcount;
#ifdef DEBUG
        cout << "final len " << dp[n].len << endl;
        cout << "mostcount " << mostcount << endl;
        cout << "mostval " << mostval << endl;
#endif
        if (mostcount * 2 <= dp[n].len) {
            write(dp[n].len);
        } else {
            write(2 * (dp[n].len - mostcount));
        }
        puts("");
    }
    return 0;
}

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#pragma GCC optimize("O2")

#include <assert.h>

#include <inttypes.h>

#include <algorithm>

#include <iostream>

#include <vector>

using int_t = int;

using std::cin;

using std::cout;

using std::endl;

const int_t LARGE = 1e6;

char inputbuf[(int(2e8) / 1024) * 1024];

char* head = inputbuf;

inline char nextchar() {

return *(head++);

}

void initinput() {

fread(inputbuf, 1024, sizeof(inputbuf) / 1024, stdin);

}

struct State {

int64_t len;

int_t mostval;

int64_t mostcount;

inline State(int64_t len = 0, int_t mostval = 0, int64_t mostcount = 0)

: len(len), mostval(mostval), mostcount(mostcount) {}

State operator+(const State& rhs) const {

State result(len + rhs.len, 0, 0);

if (mostval == rhs.mostval)

result.mostval = mostval,

result.mostcount = mostcount + rhs.mostcount;

else {

if (mostcount > rhs.mostcount) {

result.mostcount = mostcount - rhs.mostcount;

result.mostval = mostval;

} else {

result.mostcount = rhs.mostcount - mostcount;

result.mostval = rhs.mostval;

}

return result;

}

};

std::ostream& operator<<(std::ostream& os, const State& state) {

os << "State{len=" << state.len << ",mostval=" << state.mostval

<< ",mostcount=" << state.mostcount << "}";

return os;

}

struct Opt {

int_t type;

int_t* data;

int_t datalen;

int_t x1, x2;

int64_t mostcount = 0;

} opts[LARGE + 1];

State dp[LARGE + 1];

int_t n;

template <class T>

void read(T& x) {

x = 0;

char chr = nextchar();

while (chr < '0' || chr > '9')

chr = nextchar();

while (chr >= '0' && chr <= '9') {

x = x * 10 + chr - '0';

chr = nextchar();

}

assert(x >= 0);

}

template <class T>

void write(T x) {

assert(x >= 0);

if (x > 9)

write(x / 10);

putchar('0' + x % 10);

}

int main() {

// freopen("input.txt", "r", stdin);

initinput();

int_t T;

read(T);

while (T--) {

read(n);

for (int_t i = 1; i <= n; i++) {

auto& ref = opts[i];

if (ref.data) {

delete[] ref.data;

ref.data = nullptr;

}

dp[i] = State();

read(ref.type);

if (ref.type == 1) {

int_t k;

read(k);

int_t sum = 0, val = 0;

ref.data = new int_t[k + 1];

ref.datalen = k;

for (int_t i = 1; i <= k; i++) {

int_t x;

read(x);

ref.data[i] = x;

if (x == val)

sum++;

else

sum--;

if (sum < 0) {

sum *= -1, val = x;

}

// ref.seq.shrink_to_fit();

dp[i] = State(k, val, sum);

} else {

read(ref.x1);

read(ref.x2);

}

for (int_t i = 1; i <= n; i++) {

const auto& ref1 = opts[i];

if (ref1.type == 2) {

dp[i] = dp[ref1.x1] + dp[ref1.x2];

}

#ifdef DEBUG

for (int_t i = 1; i <= n; i++) {

cout << "dp " << i << " = " << dp[i] << endl;

}

#endif

int_t mostval = dp[n].mostval;

for (int_t i = 1; i <= n; i++) {

auto& ref = opts[i];

if (ref.type == 1)

ref.mostcount = std::count(ref.data + 1,

ref.data + 1 + ref.datalen, mostval);

else

ref.mostcount = opts[ref.x1].mostcount + opts[ref.x2].mostcount;

}

int64_t mostcount = opts[n].mostcount;

#ifdef DEBUG

cout << "final len " << dp[n].len << endl;

cout << "mostcount " << mostcount << endl;

cout << "mostval " << mostval << endl;

#endif

if (mostcount * 2 <= dp[n].len) {

write(dp[n].len);

} else {

write(2 * (dp[n].len - mostcount));

}

puts("");

}

return 0;

}

CFGYM102394I CCPC2019哈尔滨 I题 Interesting Permutations

2021年4月20日 by officeyutong·0评论

签到题我都不会做，我爬了。

首先检测一些明显非法的情况:

阅读全文..

CFGYM102394A CCPC2019哈尔滨A Artful Paintings

2021年4月19日 by officeyutong·0评论

一眼看过去是差分约束板子题，实际上也就是差分约束板子题。

阅读全文..

officeyutong

不止有算法竞赛

作者： officeyutong

CFGYM103415A CCPC2021广州A Math Ball

跑的比较快的多项式板子

常系数线性递推的新做法 – 计算[x^k]P(x)/Q(x)

Nginx路由匹配模拟器

CFGYM103104I-WHUPC I Sequence

CF1521C Nastia and a Hidden Permutation

CFGYM102394L CCPC2019哈尔滨 L题 LRU Algorithm

CFGYM102394E CCPC2019哈尔滨 E题 Exchanging Gifts

CFGYM102394I CCPC2019哈尔滨 I题 Interesting Permutations

CFGYM102394A CCPC2019哈尔滨A Artful Paintings