Bzoj_2179_fft Fast Fourier _ (FFT)

Describe

http://www.lydsy.com/JudgeOnline/problem.php?id=2179

Super-Large integer multiplication

Analysis

FFT template problem.

Think of numbers as a polynomial, X is 10. Then multiply by FFT, and the final rounding is good.

Attention:

1. To put each bit plus 0.5 (or less) before rounding, and then rounding down, it should be the calculation error of floating point number ...

1#include <bits/stdc++.h>2 using namespacestd;3 4 Const intmaxn=140000;5 Const DoublePi=acos (-1.0);6 intLen;7 intREV[MAXN],ANS[MAXN];8 CharSTR[MAXN];9 structcp{//plural (complex)Ten     Doubler,i; OnecpDoubler_=0.0,Doublei_=0.0): R (R_), I (I_) {} ACpoperator+ (ConstCP &x)Const{returnCP (r+x.r,i+x.i); } -Cpoperator- (ConstCP &x)Const{returnCP (r-x.r,i-x.i); } -Cpoperator* (ConstCP &x)Const{returnCP (r*x.r-i*x.i,r*x.i+i*X.R); } the }A[MAXN],B[MAXN],A[MAXN]; - voidBrcint&len) {//Binary Reverse substitution (bit-reverse-copy) -memset (rev,-1,sizeofrev); -     intk=1, l=0; +      while(K<len) k<<=1, l++; -len=K; +rev[0]=0; rev[len-1]=len-1; A      for(intI=1; i<len-1; i++){ at         if(rev[i]!=-1)Continue; -         intx=i,y=0, m=l; -          while(m--) y<<=1, y|= (x&1), x>>=1; -Rev[i]=y; rev[y]=i; -     } - } in voidDFT (CP *a,intNintFlag) {//discrete Fourier transform (discrete-fourier-transform) -      for(intI=0; i<n;i++) a[rev[i]]=A[i]; to      for(intI=0; i<n;i++) a[i]=A[i]; +      for(intm=2; m<=n;m<<=1){ -CP wn (COS (2.0*pi/m*flag), Sin (2.0*pi/m*flag)); the          for(intI=0; i<n;i+=m) { *CP W (1.0,0.0);intK=m>>1; $              for(intj=0; j<k;j++){Panax NotoginsengCP T=w*a[i+j+k], u=a[i+j]; -a[i+j]=u+T; thea[i+j+k]=u-T; +w=w*WN; A             } the         } +     } -     if(flag==-1) for(intI=0; i<n;i++) a[i].r/=N; $ } $ voidReadin (CP *a) { -scanf"%s", str); -     intL=strlen (str); the      for(intI=0; i<l;i++) a[i].r=str[l-1-i]-'0'; - }Wuyi intMain () { thescanf"%d",&len); -len=len*2-1; Wu Readin (a); Readin (b); - BRC (len); AboutDFT (A,len,1); DFT (B,len,1); $      for(intI=0; i<len;i++) a[i]=a[i]*B[i]; -DFT (a,len,-1); -      for(intI=0; i<len;i++) ans[i]=a[i].r+0.5; -      for(intI=0; i<len;i++) ans[i+1]+=ans[i]/Ten, ans[i]%=Ten; Alen++; +      while(!ans[len]&&len) len--; the      for(inti=len;i>=0; i--) printf ("%d", Ans[i]); -     return 0; $}

View Code

2179:fft Fast Fourier time limit:10 Sec Memory limit:259 MB
submit:2567 solved:1308
[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

Hintsource

Bzoj_2179_fft Fast Fourier _ (FFT)