"bzoj2179" FFT fast fourier FFT template

2016-06-01 09:34:54

Long, long time ago, I wrote it.

Today, compared to the efficiency, it seems that the handwritten plural is much faster.

Paste the Template:

1#include <iostream>2#include <cstdio>3#include <cstdlib>4#include <cstring>5#include <algorithm>6#include <cmath>7#include <queue>8#include <complex>9 #definell Long LongTen #defineN 500020 One using namespacestd; A intRead () { -   intx=0, f=1;CharCh=GetChar (); -    while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} the    while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} -   returnx*F; - } - structcd{ +   Doubleb; -CD (Doublex=0,Doubley=0) {a=x;b=y;} +Friend CDoperator+ (CD n1,cd n2) {returnCD (n1.a+n2.a,n1.b+n2.b);} AFriend CDoperator-(CD n1,cd n2) {returnCD (n1.a-n2.a,n1.b-n2.b);} atFriend CDoperator* (CD n1,cd n2) {returnCD (n1.a*n2.a-n1.b*n2.b,n1.a*n2.b+n1.b*n2.a);} - }; - intn,m,bit=0; - Const Long DoublePi=acos (-1.0); -   - voidFFT (CD *a,intNinttype) { in    for(intI=0, j=0; i<n;i++) { -     if(j>i) swap (a[i],a[j]); to     intk=N; +      while(j& (k >>=1)) j&=~K; -j|=K; the   } *    for(intI=1; i<=bit;i++){ $CD w_n (cos (2*type*pi/(1<<i)), Sin (2*type*pi/(1<<i)));Panax Notoginseng      for(intj=0;j< (1<<bit); j+= (1<<i)) { -CD W (1,0); the        for(intk=j;k<j+ (1<< (I-1)); k++){ +CD tmp=a[k],tt=w*a[k+ (1<< (I-1))]; Aa[k]=a[k]+tt; thea[k+ (1<< (I-1))]=tmp-tt; +w=w*W_n; -       } $     } $   } -   if(type<0) for(intI=0;i< (1<<bit) (i++) a[i].a= (a[i].a+0.5)/(1<<bit);  - } the   - CD Poly1[n],poly2[n];Wuyi intc[n*2]; the   - CharCh[n]; Wu intMain () { -n=read (); Aboutscanf"%s", ch+1); $    for(intI=0; i<n;i++) poly1[i]= (Double) (ch[n-i]-'0'); -scanf"%s", ch+1); -    for(intI=0; i<n;i++) poly2[i]= (Double) (ch[n-i]-'0'); -bit=1; A    while(1<<bit< (n<<1)) bit++; +n=1<<bit; theFFT (Poly1,n,1); FFT (Poly2,n,1); -    for(intI=0; i<n;i++) poly1[i]=poly1[i]*Poly2[i]; $FFT (poly1,n,-1); the   intjin=0, top=0; the    for(intI=0; i<n;i++){ thejin+= (int) (poly1[i].a+0.5); thec[++top]=jin%Ten; -Jin/=Ten; in   } the    while(top&&c[top]==0) top--; the    while(top) Putchar (c[top--]+'0'); About   return 0; the}

View Code

2179:fft Fast Fourier time limit:10 Sec Memory limit:259 MB
submit:2552 solved:1299
[Submit] [Status] [Discuss] Description gives two n-bit 10 binary integers x and y, you need to calculate x*y. Input first line a positive integer n. The second line describes a positive integer x with the number of bits N. The third line describes a positive integer y with the number of bits N. Output outputs a row, which is the result of x*y. Sample Input1
3
4

Sample Output12

Data range:
n<=60000

"bzoj2179" FFT fast fourier FFT template