P3338 [ZJOI2014]力（FFT）

P3338 [ZJOI2014]力

Fj=∑i=1j−1qi×qj(i−j)2−∑i=j+1nqi×qj(i−j)2Ej=∑i=1j−1qi(i−j)2−∑i=j+1nqi(i−j)2f(i)=qi,g(i)=1i2,f(0)=0,g(0)=0Ej=∑i=0jf(i)g(j−i)−∑i=jnf(i)g(i−j)F_j = \sum_{i = 1} ^{j - 1} \frac{q_i \times q_j}{(i - j) ^ 2} - \sum_{i = j + 1} ^{n} \frac{q_i \times q_j}{(i - j) ^ 2}\\ E_j = \sum_{i = 1} ^{j - 1} \frac{q_i}{(i - j) ^ 2} - \sum_{i = j + 1} ^{n} \frac{q_i}{(i - j) ^ 2}\\ f(i) = q_i, g(i) = \frac{1}{i ^ 2}, f(0) = 0, g(0) = 0\\ E_j = \sum_{i = 0} ^{j} f(i) g(j - i) - \sum_{i = j} ^{n} f(i) g(i - j)\\ Fj=i=1∑j−1(i−j)2qi×qj−i=j+1∑n(i−j)2qi×qjEj=i=1∑j−1(i−j)2qi−i=j+1∑n(i−j)2qif(i)=qi,g(i)=i21,f(0)=0,g(0)=0Ej=i=0∑jf(i)g(j−i)−i=j∑nf(i)g(i−j)

前项是标准的多项式卷积，后项我们另考虑后项。

∑i=jnf(i)g(i−j)\sum\limits_{i = j} ^{n} f(i) g(i - j)i=j∑nf(i)g(i−j)，T=i−jT = i - jT=i−j，i=T+ji = T + ji=T+j，原式子∑i=0n−jf(i+j)g(i)\sum\limits_{i = 0} ^{n - j} f(i + j)g(i)i=0∑n−jf(i+j)g(i)。

我们翻转f(i)f(i)f(i)得到f′(n−i)=f(i)f'(n - i) = f(i)f′(n−i)=f(i)，代入原式子∑i=0n−jf′(n−(i+j))g(i)\sum\limits_{i= 0} ^{n - j} f'(n - (i + j)) g(i)i=0∑n−jf′(n−(i+j))g(i)。

n−j=Tn - j= Tn−j=T，∑i=0Tf′(T−i)g(i)\sum\limits_{i = 0} ^{T} f'(T - i)g(i)i=0∑Tf′(T−i)g(i)，也就是一个多项式卷积了。

#include <bits/stdc++.h>using namespace std;struct Complex {double r, i;Complex(double _r = 0, double _i = 0) : r(_r), i(_i) {}
};Complex operator + (const Complex &a, const Complex &b) {return Complex(a.r + b.r, a.i + b.i);
}Complex operator - (const Complex &a, const Complex &b) {return Complex(a.r - b.r, a.i - b.i);
}Complex operator * (const Complex &a, const Complex &b) {return Complex(a.r * b.r - a.i * b.i, a.r * b.i + a.i * b.r);
}Complex operator / (const Complex &a, const Complex &b) {return Complex((a.r * b.r + a.i * b.i) / (b.r * b.r + b.i * b.i), (a.i * b.r - a.r * b.i) / (b.r * b.r + b.i * b.i));
}typedef long long ll;const int N = 1e6 + 10;int r[N];Complex a[N], b[N], c[N];void get_r(int lim) {for (int i = 0; i < lim; i++) {r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);}
}void FFT(Complex *f, int lim, int rev) {for (int i = 0; i < lim; i++) {if (i < r[i]) {swap(f[i], f[r[i]]);}}const double pi = acos(-1.0);for (int mid = 1; mid < lim; mid <<= 1) {Complex wn = Complex(cos(pi / mid), rev * sin(pi / mid));for (int len = mid << 1, cur = 0; cur < lim; cur += len) {Complex w = Complex(1, 0);for (int k = 0; k < mid; k++, w = w * wn) {Complex x = f[cur + k], y = w * f[cur + mid + k];f[cur + k] = x + y, f[cur + mid + k] = x - y;}}}if (rev == -1) {for (int i = 0; i < lim; i++) {f[i].r /= lim;}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int n, lim = 1;scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%lf", &a[i].r);b[n - i].r = a[i].r;c[i].r = 1.0 / i / i;}n <<= 1;while (lim <= n) {lim <<= 1;}get_r(lim);FFT(a, lim, 1);FFT(b, lim, 1);FFT(c, lim, 1);for (int i = 0; i < lim; i++) {a[i] = a[i] * c[i];b[i] = b[i] * c[i];}FFT(a, lim, -1);FFT(b, lim, -1);n >>= 1;for (int i = 1; i <= n; i++) {printf("%.3f\n", a[i].r - b[n - i].r);}return 0;
}

P3338 [ZJOI2014]力（FFT）相关推荐

P3338 [ZJOI2014]力 FFT + 推式子
传送门文章目录题意: 思路: 题意: 思路: 这个式子看起来很FFTFFTFFT,让我们来化简一下. 考虑EEE中直接将qiq_iqi约掉,所以Ei=∑j=1i−1qj(i−j)2−∑j=i+1 ...
P3338 [ZJOI2014]力 [FFT]
P3338[ZJOI2014]力P3338 [ZJOI2014]力P3338[ZJOI2014]力给出n个数qi,给出Fj的定义如下: Fj=∑i<jqiqj(i−j)2−∑i>jqiq ...
P3338 [ZJOI2014]力 —— FFT
题目链接:点我啊╭(╯^╰)╮ 题目大意: 解题思路: Ei=Fiqi=∑j=1i−1qj(i−j)2−∑j=i+1nqj(i−j)2E_i=\frac{F_i}{q_i} = \sum_{j=1}^ ...
P3338 [ZJOI2014]力
P3338 [ZJOI2014]力卷积 + FFT 题意思路 Code(921ms) 传送门: https://www.luogu.com.cn/problem/P3338 题意 Fj=∑i=1j ...
[卷积系列] P3338 [ZJOI2014]力
P3338 [ZJOI2014]力 Ej=Fjqj=∑i=1j−1qi(i−j)2−∑i=j+1nqi(i−j)2E_j= \frac{F_{j}}{q_{j}} =\sum_{i=1}^{j-1} ...
BZOJ.3527.[ZJOI2014]力(FFT)
题目链接 \(Descripiton\) 给出\(q[\ ]\),\[F[j]=\sum_{i<j}\frac{q_iq_j}{(i-j)^2}-\sum_{i>j}\frac{q_iq_ ...
BZOJ3527:[ZJOI2014]力(FFT)
Description 给出n个数qi,给出Fj的定义如下: 令Ei=Fi/qi,求Ei. Input 第一行一个整数n. 接下来n行每行输入一个数,第i行表示qi. n≤100000,0<qi ...
BZOJ3527: [Zjoi2014]力 [FFT]
化简之后,发现减号左边的式子是一个卷积.右边的式子,把一个函数倒序就是卷积,分别FFT,求解答案. 大佬blog: https://blog.csdn.net/kyleyoung_ymj/articl ...
bzoj3527: [Zjoi2014]力 fft
题意:求\(E_i=\sum_{j=1}^{i-1}qj/{(i-j)^2}-\sum_{j=i+1}^{n}qj/{(i-j)^2}\) 题解:构造前几个Ei,可以发现\(E_i=a_i*b_{j- ...

P3338 [ZJOI2014]力（FFT）

P3338 [ZJOI2014]力

P3338 [ZJOI2014]力（FFT）相关推荐

最新文章

热门文章