[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。䨻 $FFT$ 的板子即可。

注意输入的数有前导零...

 //It is made by Awson on 2018.1.27
 #include <set>
 #include <map>
 #include <cmath>
 #include <ctime>
 #include <queue>
 #include <stack>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <complex>
 #include <iostream>
 #include <algorithm>
 #define LL long long
 #define dob complex<double>
 #define Abs(a) ((a) < 0 ? (-(a)) : (a))
 #define Max(a, b) ((a) > (b) ? (a) : (b))
 #define Min(a, b) ((a) < (b) ? (a) : (b))
 #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
 #define writeln(x) (write(x), putchar('\n'))
 #define lowbit(x) ((x)&(-(x)))
 using namespace std;
 const int INF = ~0u>>;
 const double pi = acos(-1.0);
 const int N = 6e4*;
 void read(int &x) {
     char ch; bool flag = ;
     for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || ); ch = getchar());
     for (x = ; isdigit(ch); x = (x<<)+(x<<)+ch-, ch = getchar());
     x *= -*flag;
 }
 void write(int x) {
     if (x > ) write(x/);
     putchar(x%+);
 }

 int n, m, L, R[N+], sum[N+];
 dob a[N+], b[N+];

 int getnum() {char ch = getchar(); while (ch < '' || ch > '') ch = getchar(); return ch-; }
 void FFT(dob *A, int o) {
     for (int i = ; i < n; i++) if (i > R[i]) swap(A[i], A[R[i]]);
     for (int i = ; i < n; i <<= ) {
         dob wn(cos(pi/i), sin(pi*o/i)), x, y;
         for (int j = ; j < n; j += (i<<)) {
             dob w(, );
             for (int k = ; k < i; k++, w *= wn) {
                 x = A[j+k], y = w*A[i+j+k];
                 A[j+k] = x+y, A[i+j+k] = x-y;
             }
         }
     }
 }
 void work() {
     read(n); n--;
     for (int i = n; i >= ; i--) a[i] = getnum();
     for (int i = n; i >= ; i--) b[i] = getnum();
     m = n<<;
     for (n = ; n <= m; n <<= ) L++;
     for (int i = ; i < n; i++) R[i] = (R[i>>]>>)|((i&)<<(L-));
     FFT(a, ), FFT(b, );
     for (int i = ; i < n; i++) a[i] *= b[i];
     FFT(a, -);
     for (int i = ; i <= m; i++) sum[i] = int(a[i].real()/n+0.5);
     for (int i = ; i <= m; i++) sum[i+] += sum[i]/, sum[i] %= ;
     if (sum[m+]) m++; while (!sum[m]) m--;
     for (int i = m; i >= ; i--) write(sum[i]);
 }
 int main() {
     work();
     return ;
 }