HDU 1402 A * B Problem Plus（FFT）

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.

 

题目大意：求A * B。

思路：快速傅里叶变换的模板题，偷的模板……也不知道使用姿势对不对QAQ。

 

代码（250MS）：（Update：2014年11月16日）

 #include <cmath>
 #include <algorithm>
 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <complex>
 using namespace std;
 typedef complex<double> Complex;
 const double PI = acos(-);

 void fft_prepare(int maxn, Complex *&e) {
     e = new Complex[ * maxn - ];
     e += maxn - ;
     e[] = ;
     for (int i = ; i < maxn; i <<= )
         e[i] = Complex(cos( * PI * i / maxn), sin( * PI * i / maxn));
     for (int i = ; i < maxn; i++)
         if ((i & -i) != i) e[i] = e[i - (i & -i)] * e[i & -i];
     for (int i = ; i < maxn; i++) e[-i] = e[maxn - i];
 }
 /* f = 1: dft; f = -1: idft */
 void dft(Complex *a, int N, int f, Complex *e, int maxn) {
     int d = maxn / N * f;
     Complex x;
     for (int n = N, m; m = n / , m >= ; n = m, d *= )
         for (int i = ; i < m; i++)
             for (int j = i; j < N; j += n)
                 x = a[j] - a[j + m], a[j] += a[j + m], a[j + m] = x * e[d * i];
     for (int i = , j = ; j < N - ; j++) {
         for (int k = N / ; k > (i ^= k); k /= );
         if (j < i) swap(a[i], a[j]);
     }
 }

 const int MAXN = ;
 Complex x1[MAXN], x2[MAXN];
 char s1[MAXN / ], s2[MAXN / ];
 int sum[MAXN];

 int main() {
     Complex* e = ;
     fft_prepare(MAXN, e);
     while(scanf("%s%s",s1,s2) != EOF) {
         int n1 = strlen(s1);
         int n2 = strlen(s2);
         int n = ;
         while(n < n1 *  || n < n2 * ) n <<= ;
         for(int i = ; i < n; ++i) {
             x1[i] = i < n1 ? s1[n1 -  - i] - '' : ;
             x2[i] = i < n2 ? s2[n2 -  - i] - '' : ;
         }

         dft(x1, n, , e, MAXN);
         dft(x2, n, , e, MAXN);
         for(int i = ; i < n; ++i) x1[i] = x1[i] * x2[i];
         dft(x1, n, -, e, MAXN);
         for(int i = ; i < n; ++i) x1[i] /= n;

         for(int i = ; i < n; ++i) sum[i] = round(x1[i].real());
         for(int i = ; i < n; ++i) {
             sum[i + ] += sum[i] / ;
             sum[i] %= ;
         }

         n = n1 + n2 - ;
         while(sum[n] <=  && n > ) --n;
         for(int i = n; i >= ;i--) printf("%d", sum[i]);
         puts("");
     }
 }