[UOJ#34]多项式乘法

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
	return x * f;
}

#define maxn 400010
const double pi = acos(-1.0);
int n, m;
struct Complex {
	double a, b;
	Complex operator + (const Complex& t) {
		Complex ans;
		ans.a = a + t.a;
		ans.b = b + t.b;
		return ans;
	}
	Complex operator - (const Complex& t) {
		Complex ans;
		ans.a = a - t.a;
		ans.b = b - t.b;
		return ans;
	}
	Complex operator * (const Complex& t) {
		Complex ans;
		ans.a = a * t.a - b * t.b;
		ans.b = a * t.b + b * t.a;
		return ans;
	}
	Complex operator *= (const Complex& t) {
		*this = *this * t;
		return *this;
	}
} a[maxn], b[maxn];

int Ord[maxn];
void FFT(Complex* x, int n, int tp) {
	for(int i = 0; i < n; i++) if(i < Ord[i]) swap(x[i], x[Ord[i]]);
	for(int i = 1; i < n; i <<= 1) {
		Complex wn, w; wn.a = cos(pi / i); wn.b = (double)tp * sin(pi / i);
		for(int j = 0; j < n; j += (i << 1)) {
			w.a = 1.0; w.b = 0.0;
			for(int k = 0; k < i; k++) {
				Complex t1 = x[j+k], t2 = w * x[j+k+i];
				x[j+k] = t1 + t2;
				x[j+k+i] = t1 - t2;
				w *= wn;
			}
		}
	}
	return ;
}

int main() {
	n = read(); m = read();
	for(int i = 0; i <= n; i++) a[i].a = (double)read(), a[i].b = 0.0;
	for(int i = 0; i <= m; i++) b[i].a = (double)read(), b[i].b = 0.0;
	
	int L = 0;
	m += n; for(n = 1; n <= m; n <<= 1) L++;
	for(int i = 0; i < n; i++) Ord[i] = (Ord[i>>1] >> 1) | ((i & 1) << L - 1);
	FFT(a, n, 1); FFT(b, n, 1);
	for(int i = 0; i <= n; i++) a[i] *= b[i];
	FFT(a, n, -1);
	
	for(int i = 0; i < m; i++) printf("%d ", (int)(a[i].a / n + .5)); printf("%d\n", (int)(a[m].a / n + .5));
	
	return 0;
}