用了非递归的版本(二进制翻转) , 手写的复数 , 还是T了 , 有没有带佬帮忙看看常数大在哪里了啊qaq
这是评测记录

#include <bits/stdc++.h>

using namespace std;

const double PI = acos(-1);
const int maxn = 2345678;

struct cp {
	double x, y;
	cp() { x = y = 0; }
	cp(double X, double Y) : x(X), y(Y) {}
	cp operator + (const cp& o) const {
		return cp(x + o.x, y + o.y);
	}
	cp operator - (const cp& o) const {
		return cp(x - o.x, y - o.y);
	}
	void operator *= (const cp& o) {
		double tx = x * o.x - y * o.y;
		y = x * o.y + y * o.x;
		x = tx;
	}
	void operator /= (const int& n) {
		x /= n; y /= n;
	}
}a[maxn], b[maxn];

inline int flip(int &x) {
	int t = 1, rs = 0;
	while(t < x) t <<= 1, ++rs;
	x = t;
	return rs;
}

void dft(cp *a, int n, int w2) {
	cp t[n];
	for(int i = 0; i < n; ++i) {
		int p = 0;
		for(int cnt = 0, tmp = i; cnt < w2; ++cnt, tmp >>= 1) p <<= 1, p += tmp & 1;
		t[p] = a[i];
	}
	for(int m = 2; m <= n; m <<= 1) {
		int hf = m >> 1;
		for(int i = 0; i < n; i += m)
			for(int k = 0; k < hf; ++k) {
				cp omg = cp(cos(2*PI*k/m), sin(2*PI*k/m));
				omg *= t[i+k+hf];
				t[i+k+hf] = t[i+k] - omg;
				t[i+k] = t[i+k] + omg;
			}
	}
	for(int i = 0; i < n; ++i) a[i] = t[i];
}

void idft(cp *a, int n, int w2) {
	reverse(a + 1, a + n);
	dft(a, n, w2);
	for(int i = 0; i < n; ++i) a[i] /= n;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int n, m;
	cin >> n >> m;
	for(int i = 0; i <= n; ++i) cin >> a[i].x;
	for(int i = 0; i <= m; ++i) cin >> b[i].x;
	int len = n + m + 1;
	int w2 = flip(len);
	dft(a, len, w2);
	dft(b, len, w2);
	for(int i = 0; i < len; ++i) a[i] *= b[i];
	idft(a, len, w2);
	for(int i = 0; i <= n + m; ++i) cout << (int)(a[i].x + 0.5) << ' ';
	cout << endl;
	return 0;
}