【BZOJ 3527】【ZJOI 2014】力

代换一下变成多项式卷积，这里是的答案是两个卷积相减，FFT求一下两个卷积就可以啦

详细的题解：http://www.cnblogs.com/iwtwiioi/p/4126284.html

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 500003;
const double Pi = acos(- 1.0);
struct cp {
	double r, i;
	cp (double _r = 0.0, double _i = 0.0) : r(_r), i(_i) {}
	cp operator + (const cp &x) const {return cp(r + x.r, i + x.i);}
	cp operator - (const cp &x) const {return cp(r - x.r, i - x.i);}
	cp operator * (const cp &x) const {return cp(r * x.r - i * x.i, r * x.i + i * x.r);}
};
int rev[N];
cp A[N];
void DFT(cp *a, int n, int flag) {
	for(int i = 0; i < n; ++i) A[rev[i]] = a[i];
	for(int i = 0; i < n; ++i) a[i] = A[i];
	for(int m = 2; m <= n; m <<= 1) {
		cp wn(cos(2.0 * Pi / m * flag), sin(2.0 * Pi / m * flag));
		int mid = m >> 1;
		for(int i = 0; i < n; i += m) {
			cp w(1.0);
			for(int j = 0; j < mid; ++j) {
				cp u = a[i + j], t = a[i + j + mid] * w;
				a[i + j] = u + t;
				a[i + j + mid] = u - t;
				w = w * wn;
			}
		}
	}
	if (flag == -1)
		for(int i = 0; i < n; ++i)
			a[i].r /= n;
}
void init(int &n) {
	int k = 1, L = 0;
	for(; k < n; k <<= 1, ++L);
	n = k;
	for(int i = 0; i < n; ++i) {
		int t = i, ret = 0;
		for(int j = 0; j < L; ++j)
			ret <<= 1, ret |= (t & 1), t >>= 1;
		rev[i] = ret;
	}
}
void FFT(double *x, double *y, cp *a, cp *b, int len) {
	for(int i = 0; i < len; ++i) a[i].r = x[i], a[i].i = 0.0;
	for(int i = 0; i < len; ++i) b[i].r = y[i], b[i].i = 0.0;
	DFT(a, len, 1); DFT(b, len, 1);
	for(int i = 0; i < len; ++i) a[i] = a[i] * b[i];
	DFT(a, len, -1);
}
cp a[N], b[N];
int n, len;
double g[N], q[N], f[N], ans[N];
int main() {
	scanf("%d", &n); len = (n << 1) + 1;
	init(len);
	for(int i = 1; i <= n; ++i) scanf("%lf", &q[i]);
	for(int i = 1; i <= n; ++i) g[i] = 1.0 / i / i;
	for(int i = 0; i < n; ++i) f[i] = q[n - i];
 
	FFT(q, g, a, b, len);
	for(int i = 1; i <= n; ++i) ans[i] = a[i].r;
	FFT(f, g, a, b, len);
	for(int i = 1; i <= n; ++i) ans[i] -= a[n - i].r;
	for(int i = 1; i <= n; ++i) printf("%.3lf\n", ans[i]);
 
	return 0;
}

题面如下，BZOJ上没有题面喔：

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Description

给出n个数q_i，给出F_j的定义如下： 
 
令E_i=F_i/q_i，求E_i.

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。

Output

n行，第i行输出E_i。 
与标准答案误差不超过1e-2即可。

Sample Input

5
4006373.885184
15375036.435759
1717456.469144
8514941.004912
1410681.345880

Sample Output

-16838672.693
3439.793
7509018.566
4595686.886
10903040.872

Hint

对于30%的数据，n≤1000。 
对于50%的数据，n≤60000。 
对于100%的数据，n≤100000，0<q_i<1000000000。

Source

感谢nodgd放题