代码

#include <bits/stdc++.h>

using f64 = double;
using cpx = std::complex<f64>;
using u64 = unsigned long long;

constexpr int N = 1 << 20;
constexpr int bceil(int x) {
    if(x == 1 || x == 0){return x;}
    return 1 << (std::__lg(x - 1) + 1); 
}

namespace f_f_t
{
	const f64 Pi_2 = acos(-1.0) / 2;
	cpx w[N >> 3];
	int init_l;
	void init(int l)
	{
		if (l <= init_l)
		{
			return;
		}
		int t = std::__lg(l - 1);
		l = 1 << t, *w = cpx(1.0, 0.0), init_l = l << 1;
		for (int i = 1; i < l; i <<= 1)
		{
			w[i] = std::polar(1.0, Pi_2 / i);
		}
		for (int i = 1; i < l; ++i)
		{
			w[i] = w[i & (i - 1)] * w[i & -i];
		}
	}
	void dif(cpx *f, int L)
	{
		for (int l = L >> 1, r = L; l; l >>= 1, r >>= 1)
		{
			for (cpx *j = f, *o = w; j != f + L; j += r, ++o)
			{
				for (cpx *k = j; k != j + l; ++k)
				{
					cpx x = *k, y = k[l] * *o;
					*k = x + y, k[l] = x - y;
				}
			}
		}
	}
	void dit(cpx *f, int L)
	{
		for (int l = 1, r = 2; l < L; l <<= 1, r <<= 1)
		{
			for (cpx *j = f, *o = w; j != f + L; j += r, ++o)
			{
				for (cpx *k = j; k != j + l; ++k)
				{
					cpx x = *k, y = k[l];
					*k = x + y, k[l] = (x - y) * std::conj(*o);
				}
			}
		}
	}
	void Conv(f64 *f, int lim, f64 *g){
		cpx *F = (cpx*)f, *G = (cpx*)g;
		int l = lim >> 1;
		init(l), dif(F, l), dif(G, l);
		f64 fx = 2.0 / lim, fx2 = 0.5 / lim;
		F[0] = (F[0] * G[0] + 2 * F[0].imag() * G[0].imag()) * fx, F[1] = F[1] * G[1] * fx;
		for (int k = 2, m = 3; k < l; k <<= 1, m <<= 1)
		{
			for (int i = k, j = i + k - 1; i < m; ++i, --j)
			{
				cpx oi = (F[i] + std::conj(F[j])), hi = (F[i] - std::conj(F[j]));
                cpx Oi = (G[i] + std::conj(G[j])), Hi = (G[i] - std::conj(G[j]));
				cpx r0 = oi * Oi - hi * Hi * ((i & 1) ? -w[i >> 1] : w[i >> 1]), r1 = Oi * hi + oi * Hi;
                F[i] = (r0 + r1) * fx2, F[j] = std::conj(r0 - r1) * fx2;
			}
		}
        dit(F, l);
	}
}

f64 F[N >> 1], G[N >> 1];

char buf[N << 1];

void solve(){
	int tot = fread(buf, 1, sizeof(buf), stdin);
	char *bga = buf, *eda = buf, *bgb, *edb = buf + tot;
	while(!isdigit(*(edb - 1))){--edb;}
	{
		for(;(((*((u64*)eda))+0x5f5f5f5f5f5f5f5f)&0x8080808080808080)==0x8080808080808080;eda+=8){}
		for(;*eda>32;){++eda;}
		for(bgb = eda; !isdigit(*bgb); ++bgb){}
	}
	auto radX = [&](char* bg, char* ed, f64* out){
		char *edr = bg + 4, *pos = ed;
		for(; pos > edr; pos -= 4){
			*out++ = *(pos - 4) * 1000 + *(pos - 3) * 100 + *(pos - 2) * 10 + *(pos - 1) - 53328;
		}
		return *out++ = std::stoi(std::string(bg, pos)), out;
	};
	int n = radX(bga, eda, F) - F, m = radX(bgb, edb, G) - G, lim = std::max(8, bceil(n + m - 1));
	std::fill(F + n, F + lim, 0.0), std::fill(G + m, G + lim, 0.0), f_f_t::Conv(F, lim, G);
	{
		struct ict{
			int num[10000];
			//小端
			ict(){
				int j = 0;
				for(int e0 = (48 << 0); e0 < (58 << 0); e0 += (1 << 0)){
					for(int e1 = (48 << 8); e1 < (58 << 8); e1 += (1 << 8)){
						for(int e2 = (48 << 16); e2 < (58 << 16); e2 += (1 << 16)){
							for(int e3 = (48 << 24); e3 < (58 << 24); e3 += (1 << 24)){
								num[j] = e0 ^ e1 ^ e2 ^ e3, ++j;
							}
						}
					}
				}
			}
		}ot;
		int o = (n + m - 2), *ed = (int*)buf + o, *c = ed;
		u64 carry = 0;
		for(int p = 0; p < o; ++p){
			carry += u64(F[p] + 0.5), *--c = ot.num[carry % 10000u], carry /= 10000u;
		}
		fprintf(stdout, "%llu", carry + u64(F[o] + 0.5)), fwrite(c, 4, ed - c, stdout);
	}	
}

int main()
{
	std::cin.tie(nullptr)->sync_with_stdio(false);
	solve();
	return 0;
}

/*
然而，除了极其庞大的材料资源需要，该仪器无法在其制造者并不知晓其为何被建造的情况下加以制造——这种了解会需要知晓Θ'，而这一点已证实对计划是致命的。
*/

评测结果