鏈接：

http://uoj.ac/problem/34

代碼：

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     for (int i = 0; i <= m; i++) printf("%d ", (int)(a[i].real() + 0.5));
70     return 0;
71 }

UOJ 34 FFT