HYSBZ 3527 力

最新推荐文章于 2020-03-10 12:10:08 发布

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HYSBZ 3527 力

题目描述：

输入一个数列 $q_i \ (1 < i < N)$ ，令：

F j = \sum i < j q i q j ( i - j ) 2 - \sum i > j q i q j ( i - j ) 2

$F_j = \sum_{i < j} \frac{q_i q_j}{(i - j)^2} - \sum_{i > j} \frac{q_i q_j}{(i - j)^2}$
计算所有

Ei=Fiqi $E_i = \frac{F_i}{q_i}$ 。

题解：

打表，脑补之后发现，构造函数：

A (x) = \sum i = 0 i < N q i + 1 x i B (x) = \sum i = N i < 2 N - 2 x i ( i - N + 1 ) 2 - \sum i = 0 i < N - 1 x i ( N - i - 1 ) 2

$A(x) = \sum_{i=0}^{i < N} q_{i+1} x^i \\ B(x) = \sum_{i=N}^{i < 2 N - 2} \frac{x^i}{(i - N + 1)^2} - \sum_{i=0}^{i < N - 1} \frac{x^i}{(N - i - 1)^2}$

Ei $E_i$ 就等于

A⋅B $A \cdot B$ 的

i+N−1 $i + N - 1$ 次项系数。FFT即可。

题目链接： vjudge 原网站

代码：

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;

#define MAXN 1000010

const double pi = acos(-1);

static struct tComplex
{
    double x, y;
    tComplex(): x(0), y(0) {}
    tComplex(double a, double b): x(a), y(b) {}
    friend tComplex operator + (tComplex a, tComplex b)
    {
        return tComplex(a.x + b.x, a.y + b.y);
    }
    friend tComplex operator - (tComplex a, tComplex b)
    {
        return tComplex(a.x - b.x, a.y - b.y);
    }
    friend tComplex operator * (tComplex a, tComplex b)
    {
        return tComplex(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
    }
} A[MAXN], B[MAXN];
static int N, L, binL, R[MAXN];

inline void FFT(tComplex A[], int type)
{
    for (int i = 0; i < binL; i++)
        if (R[i] > i) swap(A[i], A[R[i]]);
    for (int i = 1; i < binL; i <<= 1)
    {
        tComplex wn = tComplex(cos(pi / i), sin(type * pi / i));
        for (int j = 0; j < binL; j += (i << 1))
        {
            tComplex w = tComplex(1, 0);
            for (int k = 0; k < i; k++, w = w * wn)
            {
                tComplex x = A[j + k], y = A[i + j + k] * w;
                A[j + k] = x + y;
                A[i + j + k] = x - y;
            }
        }
    }
    if (!~type)
        for (int i = 0; i < binL; i++) A[i].x /= (double)binL;
}

int main()
{
    scanf("%d", &N);
    N--;
    for (int i = 0; i <= N; i++) scanf("%lf", &A[i].x);
    for (int i = 0; i < N; i++) B[i].x = (-1.0) / ((long long)(N - i) * (N - i));
    for (int i = N + 1; i <= 2 * N; i++) B[i].x = -B[2 * N - i].x;
    for (L = 0, binL = 1; binL <= 4 * N; L++, binL <<= 1);
    for (int i = 0; i < binL; i++)
        R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    FFT(A, 1);
    FFT(B, 1);
    for (int i = 0; i <= binL; i++) A[i] = A[i] * B[i];
    FFT(A, -1);
    for (int i = N; i <= 2 * N; i++) printf("%.3lf\n", A[i].x);
    return 0;
}