[Luogu 1919]【模板】A*B Problem升级版（FFT快速傅里叶）

Description

给出两个n位10进制整数x和y，你需要计算x*y。

Input

第一行一个正整数n。 第二行描述一个位数为n的正整数x。 第三行描述一个位数为n的正整数y。

Output

输出一行，即x*y的结果。（注意判断前导0）

Sample Input

1
3
4

Sample Output

12

HINT

n<=60000

题解

A*B Problem。和 A+B Problem 一样简单。

1 input() and print(int(input()) * int(input()))

对于一个大数 $\overline{a_na_{n-1}\cdots a_0}$ ，显然我们可以将其记为 $N=a_0\cdot 10^0+a_1\cdot 10^1+\cdots+a_n\cdot10^n$ 。将 $10^k$ 变为形式幂级数 $x^k$ ： $N=a_0\cdot x^0+a_1\cdot x^1+\cdots+a_n\cdot x^n$ 。显然这是一个多项式。? $FFT$ 的板子即可。

注意输入的数有前导零...

 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 = 6e4*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, L, R[N+5], sum[N+5];
40 dob a[N+5], b[N+5];
41
42 int getnum() {char ch = getchar(); while (ch < ‘0‘ || ch > ‘9‘) ch = getchar(); return ch-48; }
43 void FFT(dob *A, int o) {
44     for (int i = 0; i < n; i++) if (i > R[i]) swap(A[i], A[R[i]]);
45     for (int i = 1; i < n; i <<= 1) {
46         dob wn(cos(pi/i), sin(pi*o/i)), x, y;
47         for (int j = 0; j < n; j += (i<<1)) {
48             dob w(1, 0);
49             for (int k = 0; k < i; k++, w *= wn) {
50                 x = A[j+k], y = w*A[i+j+k];
51                 A[j+k] = x+y, A[i+j+k] = x-y;
52             }
53         }
54     }
55 }
56 void work() {
57     read(n); n--;
58     for (int i = n; i >= 0; i--) a[i] = getnum();
59     for (int i = n; i >= 0; i--) b[i] = getnum();
60     m = n<<1;
61     for (n = 1; n <= m; n <<= 1) L++;
62     for (int i = 0; i < n; i++) R[i] = (R[i>>1]>>1)|((i&1)<<(L-1));
63     FFT(a, 1), FFT(b, 1);
64     for (int i = 0; i < n; i++) a[i] *= b[i];
65     FFT(a, -1);
66     for (int i = 0; i <= m; i++) sum[i] = int(a[i].real()/n+0.5);
67     for (int i = 0; i <= m; i++) sum[i+1] += sum[i]/10, sum[i] %= 10;
68     if (sum[m+1]) m++; while (!sum[m]) m--;
69     for (int i = m; i >= 0; i--) write(sum[i]);
70 }
71 int main() {
72     work();
73     return 0;
74 }

原文地址：https://www.cnblogs.com/NaVi-Awson/p/8367115.html