差分与前缀和

发表于 | 阅读数

求$k$阶前缀和，差分。

前缀和

差分

$NTT$ 一下即可。

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 4e6 + 10;
//const int mod = 1e9 + 7;
const int P = 1004535809, gi = 3;
const int mod = P;
const double pi = acos(-1.0);
int inc(int x, int y, int mo)
{
    if (x + y >= mo)
        return (x + y - mo);
    else
        return (x + y);
}

int del(int x, int y, int mo)
{
    if (x - y < 0)
        return (x - y + mo);
    else
        return (x - y);
}
int read()
{
    ll x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', x %= P, c = getchar();
    return x * f % P;
}

int qpow(int a, int x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mo;
        x >>= 1;
        a = 1ll * a * a % mo;
    }
    return res;
}
int rev[N];
void NTT(int *A, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(A[i], A[rev[i]]);
    for (int l = 1; l < n; l <<= 1)
    {
        int tt = qpow(gi, (P - 1) / (l << 1), P);
        int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
        for (int i = 0; i < n; i += (l << 1))
        {
            int omega = 1;
            for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P)
            {
                int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
                A[i + j] = inc(x, y, P);
                A[i + j + l] = del(x, y, P);
            }
        }
    }
    int invv = qpow(n, P - 2, P);
    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i] = 1ll * A[i] * invv % P;
}

void NTTX(int *a, int n, int *b, int m, int *ans)
{
    int ML = 1, bit = 0;
    while (ML < n + m)
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    NTT(a, ML, 1);
    NTT(b, ML, 1);
    for (int i = 0; i < ML; i++)
        ans[i] = 1ll * a[i] * b[i] % P;
    NTT(ans, ML, -1);
}
int a[N], g[N], ans[N];
int main()
{
    int n = read();
    int k = read();
    int op = read();
    for (int i = 1; i <= n; i++)
        a[i] = read();
    g[1] = 1;
    for (int i = 2; i <= n; i++)
    {
        if (op == 0)
            g[i] = 1ll * g[i - 1] * (k + i - 2) % mod * qpow(i - 1, mod - 2, mod) % mod;
        else
            g[i] = (1ll * g[i - 1] * (k - i + 2) % mod * qpow(i - 1, mod - 2, mod) % mod * (-1) + mod) % mod;
    }
    NTTX(a, n, g, n, ans);
    for (int i = 2; i <= n + 1; i++)
        printf("%d ", ans[i]);
}