Codeforces Round #488 by NEAR (Div. 1)

题目链接：

Nikita and Order Statistics

CF993E Nikita and Order Statistics

Nikita likes tasks on order statistics, for example, he can easily find the \(k\)-th number in increasing order on a segment of an array. But now Nikita wonders how many segments of an array there are such that a given number \(x\) is the \(k\)-th number in increasing order on this segment. In other words, you should find the number of segments of a given array such that there are exactly \(k\) numbers of this segment which are less than \(x\).

Nikita wants to get answer for this question for each \(k\) from \(0\) to \(n\), where \(n\) is the size of the array.

Input

The first line contains two integers \(n\) and \(x (1 \le n \le 2 \cdot 10^5, -10^9 \le x \le 10^9)\).

The second line contains \(n\) integers \(a_1,a_2,…,a_n (-10^9 \le a_i \le 10^9)\) — the given array.

Output

Print \(n+1\) integers, where the \(i\)-th number is the answer for Nikita's question for \(k=i−1\).

Examples

input

5 3
1 2 3 4 5

output

6 5 4 0 0 0

input

2 6
-5 9

output

1 2 0

input

6 99
-1 -1 -1 -1 -1 -1

output

0 6 5 4 3 2 1

Solution

题意

给定长度为 \(n\) 的数组，对于 \(k \in [0, n]\)，求有多少个区间满足区间内恰好有 \(k\) 个数比 \(x\) 小。

题解

FFT

首先，对于数组中的每一个数，如果比 \(x\) 小，那么把这个数置为 \(1\)，否则置为 \(0\)。然后求该数组的前缀和 \(s\)。

设 \(f[i]\) 表示前缀和等于 \(i\) 的个数。那么 \(ans[k] = \sum_{i=k}^nf[i]f[i-k] = \sum_{i=k}^nf[i]f[n-(n+k-i)] = \sum_{i+j=n+k}f[i]f[n-j]\)。

令 \(g[i] = f[n - i]\)，则 \(ans[k] = \sum_{i+j=n+k}f[i]g[j]\)。

问题就转化为求 \(f\) 与 \(g\) 的卷积。

注意 \(k = 0\) 的情况要特判。此时会有 \(n + 1\) 次自己和自己算到一起，也就是空串的情况。所以要对 \(ans[0]\) 减去 \(n + 1\)。减完后还要除以 \(2\)，因为 \(k \neq 0\) 时，由于 \(s\) 单调不减，因此 \(s[i] \ge s[i - k]\)。此时一定保证区间的左端点在前，右端点在后。而 \(k = 0\) 时，不能保证左端点一定在前，右端点一定在后，所以要除以 \(2\)。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double PI = acos(-1);
const int maxn = 1e6 + 10;
typedef complex<double> Complex;

ll n, x;
ll s[maxn], cnt[maxn];
Complex f[maxn], g[maxn];
ll ans[maxn];
int bit = 2, rev[maxn];

void get_rev(){
    memset(rev, 0, sizeof(rev));
    while(bit <= n * 2) bit <<= 1;
    for(int i = 0; i < bit; ++i) {
        rev[i] = (rev[i >> 1] >> 1) | (bit >> 1) * (i & 1);
    }
}

void FFT(Complex *a, int op) {
    for(int i = 0; i < bit; ++i) {
        if(i < rev[i]) swap(a[i], a[rev[i]]);
    }
    for(int mid = 1; mid < bit; mid <<= 1) {
        Complex wn = Complex(cos(PI / mid), op * sin(PI / mid));
        for(int j = 0; j < bit; j += mid<<1) {
            Complex w(1, 0);
            for(int k = 0; k < mid; ++k, w = w * wn) {
                Complex x = a[j + k], y = w * a[j + k + mid];
                a[j + k] = x + y, a[j + k + mid] = x - y;
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> x;
    ++cnt[0];
    for(int i = 1; i <= n; ++i) {
        int a;
        cin >> a;
        s[i] = s[i - 1] + (a < x);
        ++cnt[s[i]];
    }
    for(int i = 0; i <= n; ++i) {
        f[i] = cnt[i];
        g[n - i] = cnt[i];
    }
    get_rev();
    FFT(f, 1), FFT(g, 1);
    for(int i = 0; i < bit; ++i) {
        f[i] = f[i] * g[i];
    }
    FFT(f, -1);
    for(int i = 0; i <= n; ++i) {
        ans[i] = (ll)(f[n + i].real() / bit + 0.5);
    }
    cout << (ans[0] - n - 1) / 2 << " ";
    for(int i = 1; i <= n; ++i) {
        cout << ans[i] << " ";
    }
    cout << endl;
    return 0;
}