BZOJ 3527: [Zjoi2014]力 FFT

3527: [Zjoi2014]力

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

 

题解：

　　

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 5e5+10, M = 1e3+20,inf = 2e9,mod = 1e9+7;

struct Complex {
    double r , i ;
    Complex () {}
    Complex ( double r , double i ) : r ( r ) , i ( i ) {}
    Complex operator + ( const Complex& t ) const {
        return Complex ( r + t.r , i + t.i ) ;
    }
    Complex operator - ( const Complex& t ) const {
        return Complex ( r - t.r , i - t.i ) ;
    }
    Complex operator * ( const Complex& t ) const {
        return Complex ( r * t.r - i * t.i , r * t.i + i * t.r ) ;
    }
} ;

void FFT ( Complex y[] , int n , int rev ) {
    for ( int i = 1 , j , t , k ; i < n ; ++ i ) {
        for ( j = 0 , t = i , k = n >> 1 ; k ; k >>= 1 , t >>= 1 ) j = j << 1 | t & 1 ;
        if ( i < j ) swap ( y[i] , y[j] ) ;
    }
    for ( int s = 2 , ds = 1 ; s <= n ; ds = s , s <<= 1 ) {
        Complex wn = Complex ( cos ( rev * 2 * pi / s ) , sin ( rev * 2 * pi / s ) ) , w ( 1 , 0 ) , t ;
        for ( int k = 0 ; k < ds ; ++ k , w = w * wn ) {
            for ( int i = k ; i < n ; i += s ) {
                y[i + ds] = y[i] - ( t = w * y[i + ds] ) ;
                y[i] = y[i] + t ;
            }
        }
    }
    if ( rev == -1 ) for ( int i = 0 ; i < n ; ++ i ) y[i].r /= n ;
}

double q[N],num[N];
Complex s[N],t[N];
int n;
int main() {
    scanf("%d",&n);
    for(int i = 1; i <= n; ++i)
        scanf("%lf",&q[i]);
    for(int i = 0; i < n-1; ++i)
        num[i] = (double)-1.0/(1.0*(n-i-1)*(n-i-1));
    num[n-1] = 0;
    for(int i = n; i < 2*n-1; ++i)
        num[i] = (double)1.0/(1.0*(i-n+1)*(i-n+1));
    int n1 = 1;
    for(n1=1;n1<2*n-1;n1<<=1);

    for(int i = 0; i < 2*n-1; ++i) s[i] = Complex(num[i],0);
    for(int i = 2*n-1; i < n1; ++i) s[i] = Complex(0,0);

    for(int i = 0; i < n; ++i)t[i] = Complex(q[i+1],0);
    for(int i = n; i < n1; ++i) t[i] = Complex(0,0);

    FFT(s,n1,1);FFT(t,n1,1);

    for(int i = 0; i < n1; ++i) t[i] = t[i]*s[i];

    FFT(t,n1,-1);

    int cnt = 1;
    for(int i = n-1; i < 2*n-1; ++i) {
        printf("%.3f\n",t[i].r);
    }
    return 0;
}