FFT用于高效大数乘法（当模板用)

转载来源:https://blog.csdn.net/zj_whu/article/details/72954766

#include <cstdio>
#include <cmath>
#include <complex>
#include <cstring>

using namespace std;
const double PI(acos(-1.0));
typedef complex<double> C;

const int N = (1 << 20);
int ans[N];
C a[N], b[N];
char s[N], t[N];

void bit_reverse_swap(C* a, int n) {
  for (int i = 1, j = n >> 1, k; i < n - 1; ++i) {
    if (i < j) swap(a[i], a[j]);
    // tricky
    for (k = n >> 1; j >= k; j -= k, k >>= 1)  // inspect the highest "1"
      ;
    j += k;
  }
}

void FFT(C* a, int n, int t) {
  bit_reverse_swap(a, n);
  for (int i = 2; i <= n; i <<= 1) {
    C wi(cos(2.0 * t * PI / i), sin(2.0 * t * PI / i));
    for (int j = 0; j < n; j += i) {
      C w(1);
      for (int k = j, h = i >> 1; k < j + h; ++k) {
        C t = w * a[k + h], u = a[k];
        a[k] = u + t;
        a[k + h] = u - t;
        w *= wi;
      }
    }
  }
  if (t == -1) {
    for (int i = 0; i < n; ++i) {
      a[i] /= n;
    }
  }
}

int trans(int x) {
  return 1 << int(ceil(log(x) / log(2) - 1e-9));  // math.h/log() 以e为底
}

int main() {
  //  freopen("test0.in","r",stdin);
  // freopen("test0b.out","w",stdout);
  int n, m, l;
  for (; ~scanf("%s%s", s, t);) {
    n = strlen(s);
    m = strlen(t);
    l = trans(n + m - 1);  // n次*m次不超过n+m-1次
    for (int i = 0; i < n; ++i) a[i] = C(s[n - 1 - i] - '0');
    for (int i = n; i < l; ++i) a[i] = C(0);
    for (int i = 0; i < m; ++i) b[i] = C(t[m - 1 - i] - '0');
    for (int i = m; i < l; ++i) b[i] = C(0);

    FFT(a, l, 1);  //把A和B换成点值表达
    FFT(b, l, 1);
    for (int i = 0; i < l; ++i)  //点值做乘法
      a[i] *= b[i];
    FFT(a, l, -1);  //逆DFT
    for (int i = 0; i < l; ++i) ans[i] = (int)(a[i].real() + 0.5);
    ans[l] = 0;  // error-prone :'l' -> '1'
    for (int i = 0; i < l; ++i) {
      ans[i + 1] += ans[i] / 10;
      ans[i] %= 10;
    }
    int p = l;
    for (; p && !ans[p]; --p)
      ;
    for (; ~p; putchar(ans[p--] + '0'))
      ;
        puts("");
    }
    return 0;
}