• 快速傅里叶变换 FFT


      1 #include<iostream>
      2 #include<cstdio>
      3 #include<cmath>
      4 #include<cstring>
      5 #include<vector>
      6 #include<algorithm>
      7 using namespace std;
      8 #define MAXN 550000+100
      9 #define PI 3.14159265358
     10 #define EPS 1e-2
     11 char a[MAXN],b[MAXN];
     12 int n,m,top=0,label=0;
     13 int ans[MAXN];
     14 struct complex
     15 {
     16     double R,I;
     17     complex(){}
     18     complex(double r0,double i0): R(r0),I(i0) {}
     19 };
     20 complex operator *(const complex& A,const complex &B)
     21 {
     22     return complex(A.R*B.R-A.I*B.I,A.R*B.I+B.R*A.I);
     23 }
     24 complex operator +(const complex& A,const complex &B)
     25 {
     26     return complex(A.R+B.R,A.I+B.I);
     27 }
     28 complex operator -(const complex& A,const complex &B)
     29 {
     30     return complex(A.R-B.R,A.I-B.I);
     31 }
     32 vector<complex> C[3*MAXN];
     33 complex p[MAXN];
     34 int FFT(int first,int floor,int multi)
     35 {
     36     int i;
     37     if(n==floor) 
     38     {
     39         C[++top].push_back(p[first]);
     40         return top;
     41     }
     42     complex wn=complex(cos(2*PI/n*floor),sin(2*PI/n*floor)*multi);
     43     complex w=complex(1,0);
     44     label++;
     45     int top0=FFT(first,floor*2,multi);
     46     int top1=FFT(first+floor,floor*2,multi);
     47     top++;
     48     for(i=0;i<n/floor/2;i++,w=w*wn)
     49         C[top].push_back(C[top0][i]+w*C[top1][i]);
     50 
     51     w=complex(1,0);
     52     for(i=0;i<n/floor/2;i++,w=w*wn)
     53         C[top].push_back(C[top0][i]-w*C[top1][i]);
     54     return top;
     55 }
     56 int make_int(double x)
     57 {
     58     int A=x;
     59     if (fabs(x-A)<EPS)
     60         return A;
     61     return A+1;
     62 }
     63 void solve()
     64 {
     65     int i=1;
     66     for(i=0;i<n;i++)
     67         p[i].R=a[n-i-1]-48;
     68     while(i<n) i*=2;
     69     n=i*2;
     70     
     71     i=1;
     72     while(i<m) i*=2;
     73     n=max(i*2,n);
     74 
     75     int top0=FFT(0,1,1);
     76 
     77     memset(p,0,sizeof(p));
     78     for(i=0;i<m;i++)
     79         p[i].R=b[m-i-1]-48;
     80     int top1=FFT(0,1,1);
     81 
     82     memset(p,0,sizeof(p));
     83     for(i=0;i<n;i++)
     84         p[i]=C[top0][i]*C[top1][i];
     85     int top2=FFT(0,1,-1);
     86 
     87     for(i=0;i<n;i++) 
     88     {
     89         ans[i]+=make_int(C[top2][i].R/n);
     90         ans[i+1]=ans[i]/10;
     91         ans[i]=ans[i]%10;
     92     }
     93 
     94     for(i=n-1;i>=0;i--) if(ans[i]!=0) break;
     95     for(i;i>=0;i--) printf("%d",ans[i]);
     96     printf("\n");
     97 }
     98 
     99 int main()
    100 {
    101     int i;
    102     scanf("%s%s",a,b);
    103     n=strlen(a); m=strlen(b);
    104     solve();
    105 }
    106     
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  • 原文地址:https://www.cnblogs.com/myoi/p/2512195.html
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