FFT之大数乘法

 #include <iostream>
 #include <stdio.h>
 #include <cmath>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 using namespace std;
 #define N 50500*2
 const double PI = acos(-1.0);
 struct Vir
 {
     double re, im;
     Vir(double _re = ., double _im = .) :re(_re), im(_im){}
     Vir operator*(Vir r) { return Vir(re*r.re - im*r.im, re*r.im + im*r.re); }
     Vir operator+(Vir r) { return Vir(re + r.re, im + r.im); }
     Vir operator-(Vir r) { return Vir(re - r.re, im - r.im); }
 };
 void bit_rev(Vir *a, int loglen, int len)
 {
     for (int i = ; i < len; ++i)
     {
         int t = i, p = ;
         for (int j = ; j < loglen; ++j)
         {
             p <<= ;
             p = p | (t & );
             t >>= ;
         }
         if (p < i)
         {
             Vir temp = a[p];
             a[p] = a[i];
             a[i] = temp;
         }
     }
 }
 void FFT(Vir *a, int loglen, int len, int on)
 {
     bit_rev(a, loglen, len);

     for (int s = , m = ; s <= loglen; ++s, m <<= )
     {
         Vir wn = Vir(cos( * PI*on / m), sin( * PI*on / m));
         for (int i = ; i < len; i += m)
         {
             Vir w = Vir(1.0, );
             for (int j = ; j < m / ; ++j)
             {
                 Vir u = a[i + j];
                 Vir v = w*a[i + j + m / ];
                 a[i + j] = u + v;
                 a[i + j + m / ] = u - v;
                 w = w*wn;
             }
         }
     }
     if (on == -)
     {
         for (int i = ; i < len; ++i) a[i].re /= len, a[i].im /= len;
     }
 }
 char a[N * ], b[N * ];
 Vir pa[N * ], pb[N * ];
 int ans[N * ];
 int main()
 {
     while (scanf("%s%s", a, b) != EOF)
     {
         int lena = strlen(a);
         int lenb = strlen(b);
         int n = , loglen = ;
         while (n < lena + lenb) n <<= , loglen++;
         for (int i = , j = lena - ; i < n; ++i, --j)
             pa[i] = Vir(j >=  ? a[j] - '' : ., .);
         for (int i = , j = lenb - ; i < n; ++i, --j)
             pb[i] = Vir(j >=  ? b[j] - '' : ., .);
         for (int i = ; i <= n; ++i) ans[i] = ;

         FFT(pa, loglen, n, );
         FFT(pb, loglen, n, );
         for (int i = ; i < n; ++i)
             pa[i] = pa[i] * pb[i];
         FFT(pa, loglen, n, -);

         for (int i = ; i < n; ++i) ans[i] = pa[i].re + 0.5;
         for (int i = ; i<n; ++i) ans[i + ] += ans[i] / , ans[i] %= ;

         int pos = lena + lenb - ;
         for (; pos> && ans[pos] <= ; --pos);
         for (; pos >= ; --pos) printf("%d", ans[pos]);
         puts("");
     }
     return ;
 }