hdu----(1402)A * B Problem Plus（FFT模板）

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12665    Accepted Submission(s): 2248

Problem Description

Calculate A * B.

 

 

Input

Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

 

 

Output

For each case, output A * B in one line.

 

 

Sample Input

1 2 1000 2

 

 

Sample Output

2 2000

 

 

Author

DOOM III

快速傅里叶转换算法.....

代码：

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cmath>

 4 #include <cstring>

 5 using namespace std;

 6 #define N 50500*2

 7 const double PI=acos(-1.0);

 8 struct Vir

 9 {

10     double re,im;

11     Vir(double _re=0.,double _im=0.):re(_re),im(_im){}

12     Vir operator*(Vir r) { return Vir(re*r.re-im*r.im,re*r.im+im*r.re);}

13     Vir operator+(Vir r) { return Vir(re+r.re,im+r.im);}

14     Vir operator-(Vir r) { return Vir(re-r.re,im-r.im);}

15 };

16 void bit_rev(Vir *a,int loglen,int len)

17 {

18     for(int i=0;i<len;++i)

19     {

20         int t=i,p=0;

21         for(int j=0;j<loglen;++j)

22         {

23             p<<=1;

24             p=p|(t&1);

25             t>>=1;

26         }

27         if(p<i)

28         {

29             Vir temp=a[p];

30             a[p]=a[i];

31             a[i]=temp;

32         }

33     }

34 }

35 void FFT(Vir *a,int loglen,int len,int on)

36 {

37     bit_rev(a,loglen,len);

38 

39     for(int s=1,m=2;s<=loglen;++s,m<<=1)

40     {

41         Vir wn=Vir(cos(2*PI*on/m),sin(2*PI*on/m));

42         for(int i=0;i<len;i+=m)

43         {

44             Vir w=Vir(1.0,0);

45             for(int j=0;j<m/2;++j)

46             {

47                 Vir u=a[i+j];

48                 Vir v=w*a[i+j+m/2];

49                 a[i+j]=u+v;

50                 a[i+j+m/2]=u-v;

51                 w=w*wn;

52             }

53         }

54     }

55     if(on==-1)

56     {

57         for(int i=0;i<len;++i){

58           a[i].re/=len;

59           a[i].im/=len;

60         }

61     }

62 }

63 char a[N*2],b[N*2];

64 Vir pa[N*2],pb[N*2];

65 int ans[N*2];

66 int main ()

67 {

68     while(scanf("%s%s",a,b)!=EOF)

69     {

70         int lena=strlen(a);

71         int lenb=strlen(b);

72         int n=1,loglen=0;

73         while(n<lena+lenb) n<<=1,loglen++;

74         for(int i=0,j=lena-1;i<n;++i,--j)

75             pa[i]=Vir(j>=0?a[j]-'0':0.,0.);

76         for(int i=0,j=lenb-1;i<n;++i,--j)

77             pb[i]=Vir(j>=0?b[j]-'0':0.,0.);

78          memset(ans,0,sizeof(int)*(n+1));

79         FFT(pa,loglen,n,1);

80         FFT(pb,loglen,n,1);

81         for(int i=0;i<n;++i)

82             pa[i]=pa[i]*pb[i];

83         FFT(pa,loglen,n,-1);

84 

85         for(int i=0;i<n;++i)

86             ans[i]=pa[i].re+0.5;

87         for(int i=0;i<n;++i)

88         {

89              ans[i+1]+=ans[i]/10;

90              ans[i]%=10;

91         }

92         int pos=lena+lenb-1;

93         for(;pos>0&&ans[pos]<=0;--pos) ;

94         for(;pos>=0;--pos)

95             printf("%d",ans[pos]);

96         puts("");

97     }

98     return 0;

99 }

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