bzoj3527: [Zjoi2014]力
一句话:给出n个数qi,给出Fj的定义如下:
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![](http://oi.nks.edu.cn/images/2901.jpg)
令Ei=Fi/qi,求Ei.
思路:先把q[i]约了,然后就是:
Ei=∑j<iqj(j−i)2−∑j>iqj(j−i)2
先看左边:
令f[i]=q[i],g[i]=1/i/i
左边就是sigma f[j]*g[i-j]
然后下标和就为定值了,就是卷积了,上FFT搞一搞
右边把q反过来再上FFT搞一搞
j<i
#include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> const double pi=M_PI; const int maxn=270010; using namespace std; struct plex{ double r,i; }tmp[maxn]; plex operator +(plex a,plex b){return (plex){a.r+b.r,a.i+b.i};} plex operator -(plex a,plex b){return (plex){a.r-b.r,a.i-b.i};} plex operator *(plex a,plex b){return (plex){a.r*b.r-a.i*b.i,a.r*b.i+a.i*b.r};} int n,nn;double q[maxn],ans[maxn]; struct DFT{ plex a[maxn]; void fft(int bg,int step,int size,int op){ if (size==1) return; fft(bg,step<<1,size>>1,op),fft(bg+step,step<<1,size>>1,op); plex w=(plex){1,0},t=(plex){cos(2*pi/size),sin(2.0*pi*op/size)}; int p=bg,p0=bg,p1=bg+step; for (int i=0;i<size/2;i++){ tmp[p]=a[p0]+w*a[p1]; tmp[p+size/2*step]=a[p0]-w*a[p1]; p+=step,p0+=step*2,p1+=step*2,w=w*t; } for (int i=bg;size;i+=step,size--) a[i]=tmp[i]; } }a,b,c; int main(){ scanf("%d",&n); for (nn=1;nn<(n<<1);nn<<=1); for (int i=0;i<n;i++) scanf("%lf",&q[i]); for (int i=1;i<n;i++) b.a[i]=(plex){1.0/i/i,0}; for (int i=0;i<n;i++) a.a[i]=(plex){q[i],0}; a.fft(0,1,nn,1),b.fft(0,1,nn,1); for (int i=0;i<nn;i++) c.a[i]=a.a[i]*b.a[i]; c.fft(0,1,nn,-1); for (int i=0;i<n;i++) ans[i]+=c.a[i].r/nn; for (int i=0;i<n;i++) a.a[i]=(plex){q[n-i-1],0}; for (int i=n;i<nn;i++) a.a[i]=(plex){0,0}; a.fft(0,1,nn,1); for (int i=0;i<nn;i++) c.a[i]=a.a[i]*b.a[i]; c.fft(0,1,nn,-1); for (int i=0;i<n;i++) ans[i]-=c.a[n-i-1].r/nn; for (int i=0;i<n;i++) printf("%.7f\n",ans[i]); return 0; }
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