結果

問題 No.2506 Sum of Weighted Powers
ユーザー suisen
提出日時 2023-03-22 01:29:32
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
(最新)
AC  
(最初)
実行時間 -
コード長 4,477 bytes
コンパイル時間 2,627 ms
コンパイル使用メモリ 142,148 KB
最終ジャッジ日時 2025-02-11 16:01:22
ジャッジサーバーID
(参考情報)
judge4 / judge2
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ファイルパターン 結果
other AC * 5 WA * 37
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ソースコード

diff #
プレゼンテーションモードにする

#include <iostream>
#include <cassert>
#include <cmath>
#include <map>
long long pow(long long a, long long n, long long p) {
long long r = 1;
for (;n > 0;n >>= 1, a = a * a % p)if (n % 2 == 1)r = r * a % p;
return r;
}
int cnt(long long a, long long base, long long p) {
int ret = 0;
while (a != 1) {
a = pow(a, base, p);
++ret;
}
return ret;
}
long long inv(long long a, long long p) {
a %= p;
long long u = 1, v = 0;
long long b = p;
while (b > 0) {
long long q = a / b;
a %= b;
u -= v * q % p;
u = (u % p + p) % p;
{
u ^= v;v ^= u;u ^= v;
a ^= b;b ^= a;a ^= b;
}
}
return u < 0 ? u + p : u;
}
long long gcd(long long a, long long b) {
return a == 0 ? b : gcd(b % a, a);
}
long long peth_root(long long a, long long p, int e, long long mod) {
long long q = mod - 1;
int s = 0;
while (q % p == 0) {
q /= p;
++s;
}
long long pe = pow(p, e, mod);
long long ans = pow(a, ((pe - 1) * inv(q, pe) % pe * q + 1) / pe, mod);
long long c = 2;
while (pow(c, (mod - 1) / p, mod) == 1)
++c;
c = pow(c, q, mod);
std::map<long long, int> map;
long long add = 1;
int v = (int) std::sqrt((double) (s - e) * p) + 1;
long long mul = pow(c, v * pow(p, s - 1, mod - 1) % (mod - 1), mod);
for (int i = 0;i <= v;++i) {
map[add] = i;
add = add * mul % mod;
}
mul = inv(pow(c, pow(p, s - 1, mod - 1), mod), mod);
for (int i = e;i < s;++i) {
long long err = inv(pow(ans, pe, mod), mod) * a % mod;
long long target = pow(err, pow(p, s - 1 - i, mod - 1), mod);
for (int j = 0; j <= v; ++j) {
if (map.find(target) != map.end()) {
int x = map[target];
ans = ans * pow(c, (j + v * x) * pow(p, i - e, mod - 1) % (mod - 1), mod) % mod;
break;
}
target = target * mul % mod;
assert(j != v);
}
}
return ans;
}
long long kth_root(long long a, long long k, long long p) {
if (k > 0 && a == 0)return 0;
k %= p - 1;
long long g = gcd(k, p - 1);
if (pow(a, (p - 1) / g, p) != 1)
return -1;
a = pow(a, inv(k / g, (p - 1) / g), p);
for (long long div = 2;div * div <= g;++div) {
int sz = 0;
while (g % div == 0) {
g /= div;
++sz;
}
if (sz > 0) {
long long b = peth_root(a, div, sz, p);
a = b;
}
}
if (g > 1)
a = peth_root(a, g, 1, p);
return a;
}
#include <atcoder/modint>
using mint = atcoder::modint998244353;
namespace atcoder {
std::istream& operator>>(std::istream& in, mint &a) {
long long e; in >> e; a = e;
return in;
}
std::ostream& operator<<(std::ostream& out, const mint &a) {
out << a.val();
return out;
}
} // namespace atcoder
#include <atcoder/convolution>
mint solve(const int n, const mint x, const std::vector<mint> &a, const std::vector<mint> &b, const std::vector<mint> &c) {
if (x == 0) {
mint ans = 0;
for (int i = 0; i <= n; ++i) {
ans += a[i] * b[i] * c[0];
}
for (int i = 1; i <= n; ++i) {
ans += a[i] * b[0] * c[i];
}
return ans;
}
int cbrt_x_ = kth_root(x.val(), 3, mint::mod());
if (cbrt_x_ == -1) {
return 0;
}
const mint cbrt_x = cbrt_x_, inv_cbrt_x = cbrt_x.inv();
auto t = [&](long long k) {
return k * k * k;
};
std::vector<mint> f(n + 1), g(n + 1);
for (int i = 0; i <= n; ++i) {
const mint pow_inv_x = inv_cbrt_x.pow(t(i));
f[i] = b[i] * pow_inv_x;
g[i] = c[i] * pow_inv_x;
}
const std::vector<mint> h = atcoder::convolution(f, g);
mint ans = 0;
for (int i = 0; i <= n; ++i) {
const mint pow_x = cbrt_x.pow(t(i));
ans += a[i] * pow_x * h[i];
}
return ans;
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int n, x;
std::cin >> n >> x;
std::vector<mint> a(n + 1), b(n + 1), c(n + 1);
for (int i = 0, v; i <= n; ++i) std::cin >> v, a[i] = v;
for (int i = 0, v; i <= n; ++i) std::cin >> v, b[i] = v;
for (int i = 0, v; i <= n; ++i) std::cin >> v, c[i] = v;
std::cout << solve(n, x, a, b, c).val() << std::endl;
return 0;
}
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