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 }