CF993E Nikita and Order Statistics

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给你一个长度为 $n$ 个序列 $a$ 和一个数 $x$，对于 $0\leq k\leq n$ 求出有多少个 $a$ 的子区间满足恰好有 $k$ 个数小于 $x$。

预处理前缀和，对前缀和$NTT$没了。

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 = 1e6 + 10;
const int mod = 1e9 + 7;
int read()
{
    int 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', c = getchar();
    return x * f;
}
const double pi = acos(-1.0);
struct Complex
{
    double x, y;
    Complex(double _x = 0.0, double _y = 0.0)
    {
        x = _x;
        y = _y;
    }

    Complex operator-(const Complex &b) const
    {
        return Complex(x - b.x, y - b.y);
    }

    Complex operator+(const Complex &b) const
    {
        return Complex(x + b.x, y + b.y);
    }

    Complex operator*(const Complex &b) const
    {
        return Complex(x * b.x - y * b.y, x * b.y + y * b.x);
    }
};
int rev[N];
void FFT(Complex *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)
    {
        Complex temp(cos(pi / l), inv * sin(pi / l));
        for (int i = 0; i < n; i += (l << 1))
        {
            Complex omega(1, 0);
            for (int j = 0; j < l; j++, omega = omega * temp)
            {
                Complex x = A[i + j], y = omega * A[i + j + l];
                A[i + j] = x + y;
                A[i + j + l] = x - y;
            }
        }
    }

    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i].x = ll(A[i].x / n + 0.5);
}

void FFTX(Complex *a, int n, Complex *b, int m, Complex *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));
    FFT(a, ML, 1);
    FFT(b, ML, 1);
    for (int i = 0; i < ML; i++)
        ans[i] = a[i] * b[i];
    FFT(ans, ML, -1);
}
int a[N];
ll ans[N];
Complex f[N], g[N], c[N];
int main()
{
    int n = read(), x = read();
    for (int i = 1; i <= n; i++)
    {
        int u = read();
        a[i] = a[i - 1] + (u < x);
    }
    for (int i = 0; i <= n; i++)
    {
        if (i != n)
            f[a[i]].x++;
        if (i)
            g[a[i]].x++;
    }

    reverse(g, g + n + 1);
    FFTX(f, n, g, n, c);
    for (int k = 0; k <= n; k++)
        ans[k] = ll(c[n - k].x);
    ans[0] = 1ll * n * (n + 1) / 2;
    for (int k = 1; k <= n; k++)
        ans[0] -= ans[k];
    for (int k = 0; k <= n; k++)
        printf("%lld ", ans[k]);
}