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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P5488 差分与前缀和

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P5488 差分与前缀和

广义二项式定理：

$$
(1 + x) ^ n = \sum _ {i \ge 0} \binom n i x ^ i
$$

其中 $\binom n m = \frac {n ^ {\underline m}} {m!}$ 。

考虑前缀和的过程，每个位置对后面的所有位置贡献，即乘 $1 + x + x ^ 2 \ldots = \frac 1 {1 - x}$ 。答案为 $F * \frac 1 {(1 - x) ^ k}$ ，即 $F * \sum _ {i = 0} \binom {-k} i (-1) ^ i x ^ i = F * \sum _ {i = 0} \binom {k + i - 1} i x ^ i$

相应的，差分的答案为 $F * (1 - x) ^ k$ ，即 $F * \sum _ {i = 0} ^ k \binom k i (-1) ^ i x ^ i$ 。

查看代码 
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#include <cstdio>
#include <algorithm>
using namespace std;
template <class Type>
void read (Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        c == '-' && (flag = true);
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0';
    flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
    x < 0 && (putchar('-'), x = ~x + 1);
    x > 9 && (write(x / 10), 0);
    putchar('0' + x % 10);
}
typedef long long LL;
const int N = 5e5 + 10, mod = 1004535809;
int rev[N];
int binpow (int b, int k = mod - 2)
{
    int res = 1;
    for (; k; k >>= 1, b = (LL)b * b % mod)
        if (k & 1) res = (LL)res * b % mod;
    return res;
}
void ntt (int *x, int bit, int op)
{
    int tot = 1 << bit;
    for (int i = 0; i < tot; ++i)
        if ((rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1) > i)
            swap(x[rev[i]], x[i]);
    for (int mid = 1, t = (mod - 1) >> 1; mid < tot; mid <<= 1, t >>= 1)
    {
        int w1 = binpow(3, ~op ? t : mod - 1 - t);
        for (int i = 0; i < tot; i += mid << 1)
            for (int j = 0, k = 1; j < mid; ++j, k = (LL)k * w1 % mod)
            {
                int p = x[i | j], q = (LL)k * x[i | mid | j] % mod;
                x[i | j] = (p + q) % mod, x[i | mid | j] = (p - q) % mod;
            }
    }
    if (~op) return;
    int itot = binpow(tot);
    for (int i = 0; i < tot; ++i)
        x[i] = (LL)x[i] * itot % mod;
}
void PolyMul (int n, int *f, int m, int *g, int nm, int *w)
{
    int bit = 0;
    while (1 << bit < n + m - 1) ++bit;
    int tot = 1 << bit;
    for (int i = n; i < tot; ++i) f[i] = 0;
    for (int i = m; i < tot; ++i) g[i] = 0;
    ntt(f, bit, 1), ntt(g, bit, 1);
    for (int i = 0; i < tot; ++i)
        w[i] = (LL)f[i] * g[i] % mod;
    ntt(w, bit, -1);
    for (int i = nm; i < tot; ++i) w[i] = 0;
}
int main ()
{
    int n, k = 0, op;
    read(n);
    char c;
    while ((c = getchar()) < '0' || c > '9');
    k = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        k = (k * 10ll + c - '0') % mod;
    read(op);
    static int A[N], B[N];
    for (int i = 0; i < n; ++i) read(A[i]);
    B[0] = 1;
    for (int i = 1; i < n; ++i)
        B[i] = (LL)B[i - 1] * binpow(i) % mod * (op ? -(k - i + 1) : k + i - 1) % mod;
    PolyMul(n, A, n, B, n, A);
    for (int i = 0; i < n; ++i)
        write((A[i] + mod) % mod), putchar(' ');
    return 0;
}
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