[UOJ 34]多项式乘法

Posted 2020-10-22 NaVi_Awson

Description

这是一道模板题。

给你两个多项式，请输出乘起来后的多项式。

Input

第一行两个整数 $n$ 和 $m$，分别表示两个多项式的次数。

第二行 $n + 1$ 个整数，表示第一个多项式的 $0$ 到 $n$ 次项系数。

第三行 $m + 1$ 个整数，表示第二个多项式的 $0$ 到 $m$ 次项系数。

Output

一行 $n + m + 1$ 个整数，表示乘起来后的多项式的 $0$ 到 $n + m$ 次项系数。

Sample Input

1 2
1 2
1 2 1

Sample Output

1 4 5 2

Sample Explanation

$(1 + 2x) \cdot (1 + 2x + x^2) = 1 + 4x + 5x^2 + 2x^3$。

HINT

$0 \leq n, m \leq 10^5$，保证输入中的系数大于等于 $0$ 且小于等于 $9$。

时间限制：$1\texttt{s}$

空间限制：$256\texttt{MB}$

题解

终于会背模板 $FFT$ 辣！！

 1 //It is made by Awson on 2018.1.27
 2 #include <set>
 3 #include <map>
 4 #include <cmath>
 5 #include <ctime>
 6 #include <queue>
 7 #include <stack>
 8 #include <cstdio>
 9 #include <string>
10 #include <vector>
11 #include <cstdlib>
12 #include <cstring>
13 #include <complex>
14 #include <iostream>
15 #include <algorithm>
16 #define LL long long
17 #define dob complex<double>
18 #define Abs(a) ((a) < 0 ? (-(a)) : (a))
19 #define Max(a, b) ((a) > (b) ? (a) : (b))
20 #define Min(a, b) ((a) < (b) ? (a) : (b))
21 #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
22 #define writeln(x) (write(x), putchar(‘\n‘))
23 #define lowbit(x) ((x)&(-(x)))
24 using namespace std;
25 const int INF = ~0u>>1;
26 const double pi = acos(-1.0);
27 const int N = 1e5*4;
28 void read(int &x) {
29     char ch; bool flag = 0;
30     for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == ‘-‘)) || 1); ch = getchar());
31     for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
32     x *= 1-2*flag;
33 }
34 void write(int x) {
35     if (x > 9) write(x/10);
36     putchar(x%10+48);
37 }
38 
39 int n, m, x, L, R[N+5];
40 dob a[N+5], b[N+5];
41 
42 void FFT(dob *A, int o) {
43     for (int i = 0; i < n; i++) if (i > R[i]) swap(A[i], A[R[i]]);
44     for (int i = 1; i < n; i <<= 1) {
45         dob wn(cos(pi/i), sin(pi*o/i)), x, y;
46         for (int j = 0; j < n; j += (i<<1)) {
47             dob w(1, 0);
48             for (int k = 0; k < i; k++, w *= wn) {
49                 x = A[j+k], y = w*A[i+j+k];
50                 A[j+k] = x+y, A[i+j+k] = x-y;
51             }
52         }
53     }
54 }
55 void work() {
56     read(n), read(m);
57     for (int i = 0; i <= n; i++) read(x), a[i] = x;
58     for (int i = 0; i <= m; i++) read(x), b[i] = x;
59     m += n;
60     for (n = 1; n <= m; n <<= 1) L++;
61     for (int i = 0; i < n; i++) R[i] = (R[i>>1]>>1)|((i&1)<<(L-1));
62     FFT(a, 1), FFT(b, 1);
63     for (int i = 0; i < n; i++) a[i] *= b[i];
64     FFT(a, -1);
65     for (int i = 0; i <= m; i++) write(int(a[i].real()/n+0.5)), putchar(‘ ‘);
66 } 
67 int main() {
68     work();
69     return 0;
70 }