3527: [Zjoi2014] Force

The main topic: give the number of QI, the definition of FJ

      

Make Ei=fi/qi, Seek Ei.

Set a[i]=q[i],b[i]=1/(i^2).

Set C[i]=sigma (A[j]*b[i-j]), D[i]=sigma (A[n-j-1]*b[i-j]).

Then the e[i]=c[i]-d[i of the request].

It is not difficult to find C[i] is already the standard convolution form, with the FFT can.

For D[i], make A ' [i]=a[n-i-1], then D[i]=sigma (A[j]*b[i-j]), so also use FFT.

Code

1#include <cstdio>2#include <iostream>3#include <cmath>4#include <cstring>5#include <algorithm>6 #defineMAXN 2621447 #definePi 3.141592653589793238468 using namespacestd;9 Charch;Ten intM,n; One BOOLOK; A Long DoubleANS[MAXN]; - DoubleQ[MAXN]; - voidReadint&x) {  the      for(ok=0, Ch=getchar ();! IsDigit (CH); Ch=getchar ())if(ch=='-') ok=1;  -      for(x=0; isdigit (ch); x=x*Ten+ch-'0', ch=GetChar ());  -     if(OK) x=-x; - }  + structcomp{ -     Long DoubleRea,ima; +     voidClear () {rea=ima=0;}  ACompoperator+(ConstComp &x) {return(comp) {rea+x.rea,ima+X.ima};}  atCompoperator-(ConstComp &x) {return(comp) {rea-x.rea,ima-X.ima};}  -Compoperator*(ConstComp &x) {return(comp) {rea*x.rea-ima*x.ima,rea*x.ima+ima*X.rea};}  - }a[maxn],b[maxn],c[maxn],tmp[maxn],w,wn;  - voidFFT (Comp *a,intStintSizintStepintop) {  -     if(siz==1)return;  -FFT (a,st,siz>>1,step<<1, op), FFT (a,st+step,siz>>1,step<<1, op);  in     intx=st,x1=st,x2=st+Step; -w= (comp) {1,0},wn= (comp) {cos (op*2*pi/siz), sin (op*2*pi/siz)};  to      for(intI=0;i< (siz>>1); i++,x+=step,x1+= (step<<1), x2+= (step<<1), w=w*WN) +tmp[x]=a[x1]+ (w*a[x2]), tmp[x+ (siz>>1) *step]=a[x1]-(w*a[x2]);  -      for(inti=st;siz;i+=step,siz--) a[i]=Tmp[i]; the }  * intMain () { $Read (m), n=1; Panax Notoginseng      while(n< (m<<1)) n<<=1;  -      for(intI=0; i<m;i++) scanf ("%LF",&Q[i]);  the        +      for(intI=0; i<n;i++) a[i].clear ();  A      for(intI=0; i<n;i++) b[i].clear ();  the      for(intI=0; i<m;i++) a[i].rea=Q[i]; +      for(intI=1; i<m;i++) b[i].rea=1.0/i/i; -FFT (A,0N1,1), FFT (b,0N1,1);  $      for(intI=0; i<n;i++) c[i]=a[i]*B[i]; $FFT (c,0N1,-1);  -      for(intI=0; i<m;i++) ans[i]=c[i].rea/N; -        the      for(intI=0; i<n;i++) a[i].clear ();  -      for(intI=0; i<n;i++) b[i].clear (); Wuyi      for(intI=0; i<m;i++) a[i].rea=q[m-i-1];  the      for(intI=1; i<m;i++) b[i].rea=1.0/i/i; -FFT (A,0N1,1), FFT (b,0N1,1);  Wu      for(intI=0; i<n;i++) c[i]=a[i]*B[i]; -FFT (c,0N1,-1);  About        $      for(intI=0; i<m;i++) printf ("%.9lf\n",(Double) (ans[i]-c[m-i-1].rea/N));  -     return 0;  -}

3527: [Zjoi2014] Force