BZOJ 3527: [Zjoi2014]力

3527: [Zjoi2014]力

Time Limit: 30 Sec  Memory Limit: 256 MBSec  Special Judge
Submit: 1545  Solved: 900
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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<1000000000

 

 

Output

 n行，第i行输出Ei。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

 

Source

 

[Submit][Status][Discuss]

 

额，FFT是真的不会，卷积是真的不熟，wzq_QwQ的题解是真的好。

首先把$q_{i}$除到右边，然后上下约掉，然后，发现有规律，然后构造函数，请移步wzq_QwQ的题解吧，懒得抄了。

 

#include <cmath>
#include <cstdio>
#include <iostream>

const int siz = 600005;
const double pi = acos(-1);

inline double sqr(double x) {
    return x * x;
}

int n, rev[siz];
double num[siz];

struct complex
{
    double r, i;

    inline complex(double a = 0, double b = 0)
        : r(a), i(b) {};
    inline complex operator + (const complex &a) {
        return complex(r + a.r, i + a.i);
    }
    inline complex operator - (const complex &a) {
        return complex(r - a.r, i - a.i);
    }
    inline complex operator * (const complex &a) {
        return complex(r*a.r - i*a.i, r*a.i + i*a.r);
    }
}a[siz], b[siz];

inline void FFT(complex *s, int k)
{
    for (int i = 0; i < n; ++i)
        if (i < rev[i])
            std::swap(s[i], s[rev[i]]);
    
    for (int h = 2; h <= n; h <<= 1)
    {
        complex wn(cos(2*pi*k/h), sin(2*pi*k/h));
        for (int i = 0; i < n; i += h)
        {
            complex w(1, 0);
            for (int j = 0; j < (h >> 1); ++j, w = w*wn)
            {
                complex t = s[i + j + (h >> 1)]*w;
                s[i + j + (h >> 1)] = s[i + j] - t;
                s[i + j] = s[i + j] + t;
            }
        }
    }
    
    if (k == -1)
        for (int i = 0; i < n; ++i)
            s[i].r = s[i].r / n;
}

signed main(void)
{
    scanf("%d", &n); --n;
    
    for (int i = 0; i <= n; ++i)
        scanf("%lf", &a[i].r);
        
    for (int i = 0; i < n; ++i)
        b[i].r = -1.0 / sqr(n - i);
        
    for (int i = n + 1; i <= 2*n; ++i)
        b[i].r = -1.0 * b[2*n - i].r;
        
    int m = n, k = n << 2, len = 0;
    
    for (n = 1; n <= k; n <<= 1)++len;
    
    for (int i = 0; i < n; ++i)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 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 = m; i <= 2*m; ++i)
        printf("%lf\n", a[i].r);
}

 

@Author: YouSiki