[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。

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 $N=a_0cdot 10^0+a_1cdot 10^1+cdots+a_ncdot10^n$ 。将 $10^k$ 变为形式幂级数 $x^k$ ： $N=a_0cdot x^0+a_1cdot x^1+cdots+a_ncdot x^n$ 。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('
'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int INF = ~0u>>1;
const double pi = acos(-1.0);
const int N = 6e4*4;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void write(int x) {
    if (x > 9) write(x/10);
    putchar(x%10+48);
}

int n, m, L, R[N+5], sum[N+5];
dob a[N+5], b[N+5];

int getnum() {char ch = getchar(); while (ch < '0' || ch > '9') ch = getchar(); return ch-48; }
void FFT(dob *A, int o) {
    for (int i = 0; i < n; i++) if (i > R[i]) swap(A[i], A[R[i]]);
    for (int i = 1; i < n; i <<= 1) {
        dob wn(cos(pi/i), sin(pi*o/i)), x, y;
        for (int j = 0; j < n; j += (i<<1)) {
            dob w(1, 0);
            for (int k = 0; 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 >= 0; i--) a[i] = getnum();
    for (int i = n; i >= 0; i--) b[i] = getnum();
    m = n<<1;
    for (n = 1; n <= m; n <<= 1) L++;
    for (int i = 0; i < n; i++) R[i] = (R[i>>1]>>1)|((i&1)<<(L-1));
    FFT(a, 1), FFT(b, 1);
    for (int i = 0; i < n; i++) a[i] *= b[i];
    FFT(a, -1);
    for (int i = 0; i <= m; i++) sum[i] = int(a[i].real()/n+0.5);
    for (int i = 0; i <= m; i++) sum[i+1] += sum[i]/10, sum[i] %= 10;
    if (sum[m+1]) m++; while (!sum[m]) m--;
    for (int i = m; i >= 0; i--) write(sum[i]);
}
int main() {
    work();
    return 0;
}