FFT (Fast Fourier transform): HDU 1402 A * B problem Plus

Calculate A * B.
Input each line would contain the integers A and B. Process to end of file.

Note:the length of each integer would not exceed 50000.
Output for each case, output A * B in one line.
Sample Input

1210002

Sample Output

22000
　　
Alas, the template problem, the membrane of the Bingbin template.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <algorithm>6 using namespacestd;7 8 Const DoublePI = ACOs (-1.0);9 Ten structcomplex{ One     Doubler,i; AComplexDoubler_=0.0,Doublei_=0.0) -     { -R=r_;i=I_; the     } -Complexoperator+(ConstComplex &b) -     { -         returnComplex (r+b.r,i+B.I); +     } -Complexoperator-(ConstComplex &b) +     { A         returnComplex (r-b.r,i-B.I); at     } -Complexoperator*(ConstComplex &b) -     { -         returnComplex (r*b.r-i*b.i,i*b.r+r*B.I); -     } - }; in   - voidRader (Complex *a,intlen) to { +     intK; -      for(intI=1, j=len/2; i<len-1; i++) the     { *         if(i<j) Swap (A[i],a[j]); $k=len/2;Panax Notoginseng          while(j>=k) -         { thej-=K; +k>>=1; A         } thej+=K; +     } - } $  $ voidFFT (Complex *a,intLenintOn ) - { - rader (a,len); the      for(intH=2; h<=len;h<<=1) -     {WuyiComplex WN (cos (-on*2*pi/h), sin (-on*2*pi/h)); the          for(intj=0; j<len;j+=h) -         { WuComplex W (1,0); -              for(intk=j;k<j+h/2; k++) About             { $Complex u=A[k]; -Complex v=a[k+h/2]*W; -a[k]=u+v; -a[k+h/2]=u-v; Aw=w*WN; +             } the         }  -     } $     if(on==-1) the          for(intI=0; i<len;i++) theA[i].r/=Len; the } the  - Const intMAXN =200010; in complex ARRAY1[MAXN],ARRAY2[MAXN]; the CharSTR1[MAXN],STR2[MAXN]; the intSum[maxn],len,len1,len2; About  the intMain () the { the      while(~SCANF ("%s%s", STR1,STR2)) +     { -len1=strlen (str1); theLen2=strlen (STR2); Bayilen=1; the          while(len<len1*2|| len<len2*2) len<<=1; the          for(intI=0; i<len1;i++) -Array1[i]=complex (str1[len1-i-1]-'0',0); -          for(intI=0; i<len2;i++) theArray2[i]=complex (str2[len2-i-1]-'0',0); the          the          for(inti=len1;i<len;i++) theArray1[i]=complex (0,0); -          for(inti=len2;i<len;i++) theArray2[i]=complex (0,0); the          theFFT (Array1,len,1);94FFT (Array2,len,1); the          for(intI=0; i<len;i++) thearray1[i]=array1[i]*Array2[i]; theFFT (array1,len,-1);98memset (SUM,0,sizeof(sum));  About          for(intI=0; i<len;i++){ -sum[i]+= (int) (array1[i].r+0.5);101sum[i+1]+=sum[i]/Ten;102sum[i]%=Ten;103         }104         intp=Len; the          while(!sum[p]&&p) p--;106          for(;p!=-1;p--)107printf"%d", Sum[p]);108printf"\ n"); 109     } the     return 0;111}

FFT (Fast Fourier transform): HDU 1402 A * B problem Plus