链接：
http://uoj.ac/problem/34
代码：
31 #include <complex>
32 typedef complex<double> E;
33 E a[MAXN], b[MAXN];
34 int n, m;
35
36 namespace FFT {
37 const double Pi = acos(-1);
38 int rev[MAXN], L;
39 void DFT(E *a, int f) {
40 for (int i = 0; i < n; i++) if (i < rev[i]) swap(a[i], a[rev[i]]);
41 for (int i = 1; i < n; i <<= 1) {
42 E wn(cos(Pi / i), f*sin(Pi / i));
43 for (int p = i << 1, j = 0; j < n; j += p) {
44 E w(1, 0);
45 for (int k = 0; k < i; k++, w *= wn) {
46 E x = a[j + k], y = w*a[j + k + i];
47 a[j + k] = x + y; a[j + k + i] = x - y;
48 }
49 }
50 }
51 if (f == -1) for (int i = 0; i < n; i++) a[i] /= n;
52 }
53 void main() {
54 m = n + m;
55 for (n = 1; n <= m; n <<= 1) L++;
56 for (int i = 0; i < n; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (L - 1));
57 DFT(a, 1); DFT(b, 1);
58 for (int i = 0; i < n; i++) a[i] = a[i] * b[i];
59 DFT(a, -1);
60 }
61 }
62
63 int main() {
64 cin >> n >> m;
65 int x;
66 rep(i, 0, n + 1) scanf("%d", &x), a[i] = x;
67 rep(i, 0, m + 1) scanf("%d", &x), b[i] = x;
68 FFT::main();
69 rep(i, 0, m + 1) printf("%d ", (int)(a[i].real() + 0.5));
70 return 0;
71 }