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<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

Solution

不会markdown/LaTeX的下场就是只能像我一样在word里编辑好然后粘图

Code

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<cmath>
 5 #define N (400000+100)
 6 using namespace std;
 7 
 8 double pi=acos(-1.0);
 9 double q[N],g[N],ans[N];
10 int n,fn,l,r[N];
11 struct complex
12 {
13     double x,y;
14     complex(double xx=0,double yy=0)
15     {
16         x=xx; y=yy;
17     }
18 }a[N],b[N];
19 
20 complex operator + (complex a,complex b){return complex(a.x+b.x,a.y+b.y);}
21 complex operator - (complex a,complex b){return complex(a.x-b.x,a.y-b.y);}
22 complex operator * (complex a,complex b){return complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
23 complex operator / (complex a,double b){return complex(a.x/b,a.y/b);}
24 
25 void FFT(int n,complex *a,int opt)
26 {
27     for (int i=0; i<n; ++i)
28         if (i<r[i])
29             swap(a[i],a[r[i]]);
30     for (int k=1; k<n; k<<=1)
31     {
32         complex wn=complex(cos(pi/k),opt*sin(pi/k));
33         for (int i=0; i<n; i+=(k<<1))
34         {
35             complex w=complex(1,0);
36             for (int j=0; j<k; ++j,w=w*wn)
37             {
38                 complex x=a[i+j], y=w*a[i+j+k];
39                 a[i+j]=x+y; a[i+j+k]=x-y;
40             }
41         }
42     }
43     if (opt==-1) for (int i=0; i<n; ++i) a[i]=a[i]/n;
44 }
45 
46 void Calc()
47 {
48     memset(a,0,sizeof(a)); memset(b,0,sizeof(b));
49     for (int i=0; i<=n; ++i) a[i].x=q[i];
50     for (int i=0; i<=n; ++i) b[i].x=g[i];
51     FFT(fn,a,1); FFT(fn,b,1);
52     for (int i=0; i<=fn; ++i)
53         a[i]=a[i]*b[i];
54     FFT(fn,a,-1);
55 }
56 
57 int main()
58 {
59     scanf("%d",&n);
60     for (int i=1; i<=n; ++i)
61         scanf("%lf",&q[i]);
62     for (int i=1; i<=n; ++i)
63         g[i]=1.0/i/i;
64     fn=1;
65     while (fn<=n+n) fn<<=1, l++;
66     for (int i=0; i<fn; ++i) 
67         r[i]=(r[i>>1]>>1) | ((i&1)<<(l-1));
68         
69     Calc();
70     for (int i=1; i<=n; ++i)
71         ans[i]=a[i].x;
72     for (int i=1; i<=n/2; ++i)
73         swap(q[i],q[n-i+1]);
74     Calc();
75     for (int i=1; i<=n; ++i)
76         printf("%.6lf
",ans[i]-a[n-i+1].x);
77 }