【BZOJ3527】[ZJOI3527]力

【BZOJ3527】[ZJOI2014]力

题面

bzoj

洛谷

题解

易得

\[E_i=\sum_{j<i}\frac{q_j}{(i-j)^2}-\sum_{j>i}\frac{q_j}{(i-j)^2}
\]

设\(f_i=q_i\)，\(g_i=i^2\)

\[E_i=\sum_{j<i}f_jg_{i-j}-\sum_{j>i}f_jg_{i-j}
\]

将\(f\)翻转得到\(h\)

\[E_i=\sum_{j<i}f_jg_{i-j}-\sum_{j<i}h_jg_{i-j}
\]

这™就是两个卷积啊。。。

就分别\(FFT\)求出两个卷积然后一减即可

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <complex>
using namespace std;
#define sqr(x) (1.0 * (x) * (x))
typedef complex<double> Complex;
const double PI = acos(-1.0);
const int MAX_N = 3e5 + 5;
int n, N, M, P, r[MAX_N];
double q[MAX_N], ans[MAX_N];
Complex a[MAX_N], b[MAX_N];
void FFT(Complex *p, int op) {
	for (int i = 0; i < N; i++) if (i < r[i]) swap(p[i], p[r[i]]);
	for (int i = 1; i < N; i <<= 1) {
		Complex rot(cos(PI / i), op * sin(PI / i));
		for (int j = 0; j < N; j += (i << 1)) {
			Complex w(1, 0);
			for (int k = 0; k < i; k++, w *= rot) {
				Complex x = p[j + k], y = w * p[i + j + k];
				p[j + k] = x + y, p[i + j + k] = x - y;
			}
		}
	}
}

int main () {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%lf", &q[i]);
	N = M = n - 1;
	for (int i = 0; i <= N; i++) a[i] = q[i + 1], b[i] = 1.0 / sqr(i + 1);
	M += N;
	for (N = 1; N <= M; N <<= 1, ++P) ;
	for (int i = 0; i < N; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (P - 1));
	FFT(a, 1); FFT(b, 1);
	for (int i = 0; i <= N; i++) a[i] = a[i] * b[i];
	FFT(a, -1);
	for (int i = 2; i <= n; i++) ans[i] += (double)(a[i - 2].real() / N);
	for (int i = 0; i <= N; i++) a[i].real() = b[i].real() = a[i].imag() = b[i].imag() = 0;
	for (int i = 0; i < n; i++) a[i] = q[n - i], b[i] = 1.0 / sqr(i + 1);
	FFT(a, 1); FFT(b, 1);
	for (int i = 0; i <= N; i++) a[i] = a[i] * b[i];
	FFT(a, -1);
	for (int i = n - 1; i; i--) ans[i] -= (double)(a[n - i - 1].real() / N);
	for (int i = 1; i <= n; i++) printf("%0.3lf\n", ans[i]);
	return 0;
}