"BZOJ2179" FFT fast fourier high accuracy multiply template problem

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#include <stdio.h>int main(){    puts("转载请注明出处[vmurder]谢谢");    puts("网址：blog.csdn.net/vmurder/article/details/44015679");}

Exercises

In fact, no matter, just put a template + understanding comments

Code:

#include <cmath>#include <cstdio>#include <cstring>#include <complex>#include <iostream>#include <algorithm>using namespace STD;#define N 131075intN,c[n];Complex<Double> A[n],b[n],p[n];Const DoublePi=ACOs(-1);voidFFT (Complex<Double> x[],intNintType) {intI,j,k,t;/* Press bits to flip: for example 01234567 04261537 000 000 001 100 010 010 011 110 100 00 1 101 101 011 111 111 * *     for(i=0, t=0; i<n;i++)//Bitwise ROLLOVER{if(i>t) Swap (x[i],x[t]); for(j=n>>1;(t^=j) <j;j>>=1); }//For WN    //type== 1 O'Clock forward rotation unit angle on complex plane    //Type==-1 negative rotation unit angle on complex plane     for(k=2; k<=n;k<<=1)     {Complex<Double> Wn (Cos(type*2*pi/k),Sin(type*2*pi/k)); for(i=0; i<n;i+=k) {Complex<Double> W (1,0), T;//You can write W = 1 directly, the x axis is the real part, and the Y axis is the imaginary part.              for(j=0;j<k>>1; w*=wn,j++) t=w*x[i+j+ (k>>1)], x[i+j+ (k>>1)]=x[i+j]-t, x[i+j]+=t; }    }}intMain () {intI,j,k;Cin>>n; for(GetChar (), i=n-1; i>=0; i--) A[i]=getchar ()-' 0 '; for(GetChar (), i=n-1; i>=0; i--) B[i]=getchar ()-' 0 '; for(j=n,i=1;i>>2<j;i<<=1) N=i; FFT (A,n,1), FFT (B,n,1);//Convert past     for(i=0; i<n;i++) P[i]=a[i]*b[i]; FFT (p,n,-1);//Convert back    intlen=0;//Output     for(i=0; i<n;i++) c[i]=p[i].real ()/n+0.1;//eps=0.1     for(i=0; i<n;i++)if(C[i]) len=i, c[i+1]+=c[i]/Ten, c[i]%=Ten; for(i=len;i>=0; i--)printf("%d", C[i]);return 0;}

"BZOJ2179" FFT fast Fourier high-precision multiply template problem