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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#468390#3799. It's Surely ComplexPetroTarnavskyiWA 2ms5668kbC++206.2kb2024-07-08 20:35:442024-07-08 20:35:45

Judging History

你现在查看的是最新测评结果

  • [2024-07-08 20:35:45]
  • 评测
  • 测评结果:WA
  • 用时:2ms
  • 内存:5668kb
  • [2024-07-08 20:35:44]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define RFOR(i, a, b) for(int i = (a) - 1; i >= (b); i--)
#define SZ(a) int(a.size())
#define ALL(a) a.begin(), a.end()
#define PB push_back
#define MP make_pair
#define F first
#define S second

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef __float128 db;

typedef unsigned long long ULL;

int p;
ULL mod = 9223372036737335297;
const int LEN = 1 << 21;

ULL add(ULL a, ULL b)
{
	return a + b < mod ? a + b : a + b - mod;
}

ULL sub(ULL a, ULL b)
{
	return a >= b ? a - b : a + mod - b;
}

ULL mult(ULL a, ULL b)
{
	return (__int128)a * b % mod;
}

ULL binpow(ULL a, ULL n)
{
	ULL res = 1;
	while (n)
	{
		if (n & 1)
			res = mult(res, a);
		a = mult(a, a);
		n /= 2;
	}
	return res;
}

const ULL GEN = binpow(3, (mod - 1) / LEN);
const ULL IGEN = binpow(GEN, mod - 2);

void fft(vector<ULL>& a, bool inv)
{
	int lg = __builtin_ctz(SZ(a));
	FOR(i, 0, SZ(a))
	{
		int k = 0;
		FOR(j, 0, lg)
			k |= ((i >> j) & 1) << (lg - j - 1);
		if (i < k)
			swap(a[i], a[k]);
	}
	for (int len = 2; len <= SZ(a); len *= 2)
	{
		ULL ml = binpow(inv ? IGEN : GEN, LEN / len);
		for (int i = 0; i < SZ(a); i += len)
		{
			ULL pw = 1;
			FOR(j, 0, len / 2)
			{
				ULL v = a[i + j];
				ULL u = mult(a[i + j + len / 2], pw);
				a[i + j] = add(v, u);
				a[i + j + len / 2] = sub(v, u);
				pw = mult(pw, ml);
			}
		}
	}
	if (inv)
	{
		ULL m = binpow(SZ(a), mod - 2);
		FOR(i, 0, SZ(a))
			a[i] = mult(a[i], m);
	}
}

VI mult(VI a, VI b)
{
	vector<ULL> A(ALL(a)), B(ALL(b));
	FOR(i, 0, SZ(a))
		assert(a[i] >= 0 && a[i] < p);
	FOR(i, 0, SZ(b))
		assert(b[i] >= 0 && b[i] < p);
	int sz = 0;
	int sum = SZ(a) + SZ(b) - 1;
	assert(sum <= 2 * p + 47);
	while ((1 << sz) < sum)
		sz++;
	A.resize(1 << sz);
	B.resize(1 << sz);
	fft(A, false);
	fft(B, false);
	FOR(i, 0, SZ(A))
		A[i] = mult(A[i], B[i]);
	fft(A, true);
	VI res(sum);
	FOR(i, 0, sum)
	{
		res[i] = A[i] % p;
		assert(res[i] >= 0);
	}
	return res;
}

int sub(int a, int b)
{
	return a - b >= 0 ? a - b : a - b + p;
}

void updAdd(int& a, int b)
{
	a += b;
	if (a >= p)
		a -= p;
}

void updSub(int& a, int b)
{
	a -= b;
	if (a < 0)
		a += p;
}

int mult(int a, int b)
{
	return (LL)a * b % p;
}

int binpow(int a, int n)
{
	int res = 1;
	while (n)
	{
		if (n & 1)
			res = mult(res, a);
		a = mult(a, a);
		n /= 2;
	}
	return res;
}

VI inverse(const VI& a, int k)
{
	assert(SZ(a) == k && a[0] != 0);
	if (k == 1)
		return {binpow(a[0], p - 2)};
	VI ra = a;
	FOR(i, 0, SZ(ra))
		if (i & 1)
			ra[i] = sub(0, ra[i]);
	
	int nk = (k + 1) / 2;
	VI t = mult(a, ra);
	t.resize(k);
	
	FOR(i, 0, nk)
		t[i] = t[2 * i];
	
	t.resize(nk);
	t = inverse(t, nk);
	t.resize(k);
	RFOR(i, nk, 1)
	{
		t[2 * i] = t[i];
		t[i] = 0;
	}
	
	VI res = mult(ra, t);
	res.resize(k);
	return res;
}

void removeLeadingZeros(VI& a)
{
	while (SZ(a) > 0 && a.back() == 0)
		a.pop_back();
}

pair<VI, VI> modulo(VI a, VI b)
{
	removeLeadingZeros(a);
	removeLeadingZeros(b);
	
	if (a.empty())
	{
		return {{}, {}};
	}
	assert(SZ(a) != 0 && SZ(b) != 0);
	int n = SZ(a), m = SZ(b);
	if (m > n)
		return MP(VI{}, a);
	
	reverse(ALL(a));
	reverse(ALL(b));
	
	VI d = b;
	d.resize(n - m + 1);
	d = mult(a, inverse(d, n - m + 1));
	d.resize(n - m + 1);
	
	reverse(ALL(a));
	reverse(ALL(b));
	reverse(ALL(d));
	
	VI res = mult(b, d);
	res.resize(SZ(a));
	FOR(i, 0, SZ(a))
		res[i] = sub(a[i], res[i]);
	
	removeLeadingZeros(d);
	removeLeadingZeros(res);
	return MP(d, res);
}

int x[LEN];
VI P[2 * LEN];

void build(int v, int tl, int tr)
{
	if (tl + 1 == tr)
	{
		P[v] = {sub(0, x[tl]), 1};
		return;
	}
	int tm = (tl + tr) / 2;
	build(2 * v + 1, tl, tm);
	build(2 * v + 2, tm, tr);
	P[v] = mult(P[2 * v + 1], P[2 * v + 2]);
}
int Ans[LEN];
void solve(int v, int tl, int tr, const VI& q)
{
	if (SZ(q) == 0)
		return;
	if (tl + 1 == tr)
	{
		Ans[tl] = q[0];
		return;
	}
	int tm = (tl + tr) / 2;
	solve(2 * v + 1, tl, tm,
		modulo(q, P[2 * v + 1]).S);
	solve(2 * v + 2, tm, tr,
		modulo(q, P[2 * v + 2]).S);
}


complex<LL> mult(complex<LL> a, complex<LL> b)
{
	a *= b;
	a = {(a.real() % p + p) % p, (a.imag() % p + p) % p};
	return a;
}

complex<LL> binpow(complex<LL> a, LL n)
{
	complex<LL> res = {1, 0};
	while (n)
	{
		if (n & 1)
			res = mult(res, a);
		a = mult(a, a);
		n /= 2;
	}
	return res;
}

pair<VI, VI> splitRealImag(int k, const VI& a)
{
	assert(SZ(a) == k + 1);
	VI resReal(k + 1), resImag(k + 1);
	FOR(i, 0, k + 1)
	{
		int r = (k - i) % 4;
		if (r == 0)
		{
			updAdd(resReal[i], a[i]);
		}
		else if (r == 1)
		{
			updAdd(resImag[i], a[i]);
		}
		else if (r == 2)
		{
			updSub(resReal[i], a[i]);
		}
		else
		{
			updSub(resImag[i], a[i]);
		}
	}
	return {resReal, resImag};
}

VI multipointEval(const VI& a, const VI& xs)
{
	VI res;
	for (int xi : xs)
	{
		int y = 0, Pw = 1;
		for (int ai : a)
		{
			updAdd(y, mult(ai, Pw));
			Pw = mult(Pw, xi);
		}
		res.PB(y);
	}
	return res;
}

vector<complex<LL>> f(int k)
{
	FOR(i, 0, k)
		x[i] = sub(0, i);
	build(0, 0, k);
	VI pol = P[0];
	auto [polReal, polImag] = splitRealImag(k, pol);
	
	FOR(i, 0, p)
		x[i] = i;
	build(0, 0, p);
	solve(0, 0, p, modulo(polReal, P[0]).S);
	VI ysReal(p);
	FOR(i, 0, p)
		ysReal[i] = Ans[i];
	solve(0, 0, p, modulo(polImag, P[0]).S);
	VI ysImag(p);
	FOR(i, 0, p)
		ysImag[i] = Ans[i];
	vector<complex<LL>> res(p);
	FOR(j, 0, p)
		res[j] = {ysReal[j], ysImag[j]};
	return res;
}

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	LL n;
	cin >> p >> n;
	vector<complex<LL>> f1 = f(p), f2 = f(n % p + 1);
	complex<LL> ans = {1, 0};
	FOR(j, 1, p)
		ans = mult(ans, complex<LL>(0, j));
	ans = binpow(ans, n / p);
	FOR(j, 1, n % p + 1)
	{
		ans = mult(ans, complex<LL>(0, j));
	}
	ans = binpow(ans, n / p + 1);
	FOR(xi, 1, p)
	{
		if (xi > n)
			break;
		ans = mult(ans, binpow(mult(binpow(f1[xi], n / p), f2[xi]), (n - xi) / p + 1));
	}
	cout << ans.real() << " " << ans.imag() << "\n";
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 2ms
memory: 5668kb

input:

2 1

output:

1 1

result:

ok single line: '1 1'

Test #2:

score: -100
Wrong Answer
time: 2ms
memory: 5632kb

input:

2 2

output:

0 0

result:

wrong answer 1st lines differ - expected: '1 1', found: '0 0'