結果

問題 No.2506 Sum of Weighted Powers
ユーザー suisensuisen
提出日時 2023-03-22 02:47:06
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 4,523 bytes
コンパイル時間 2,784 ms
コンパイル使用メモリ 147,728 KB
実行使用メモリ 11,988 KB
最終ジャッジ日時 2023-10-24 06:58:42
合計ジャッジ時間 5,400 ms
ジャッジサーバーID
(参考情報)
judge15 / judge14
このコードへのチャレンジ
(要ログイン)

テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
4,348 KB
testcase_01 AC 2 ms
4,348 KB
testcase_02 AC 2 ms
4,348 KB
testcase_03 WA -
testcase_04 WA -
testcase_05 WA -
testcase_06 AC 3 ms
4,348 KB
testcase_07 WA -
testcase_08 WA -
testcase_09 AC 3 ms
4,348 KB
testcase_10 WA -
testcase_11 WA -
testcase_12 AC 1 ms
4,348 KB
testcase_13 AC 2 ms
4,348 KB
testcase_14 WA -
testcase_15 AC 2 ms
4,348 KB
testcase_16 WA -
testcase_17 WA -
testcase_18 WA -
testcase_19 WA -
testcase_20 AC 2 ms
4,348 KB
testcase_21 WA -
testcase_22 WA -
testcase_23 WA -
testcase_24 WA -
testcase_25 WA -
testcase_26 AC 184 ms
11,268 KB
testcase_27 WA -
testcase_28 WA -
testcase_29 WA -
testcase_30 WA -
testcase_31 AC 214 ms
11,988 KB
testcase_32 WA -
testcase_33 WA -
testcase_34 WA -
testcase_35 AC 49 ms
4,824 KB
testcase_36 AC 56 ms
5,088 KB
testcase_37 AC 2 ms
4,348 KB
testcase_38 WA -
testcase_39 WA -
testcase_40 WA -
testcase_41 AC 2 ms
4,348 KB
権限があれば一括ダウンロードができます

ソースコード

diff #

#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::static_modint<943718401>;

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) {
        std::cout << -1 << std::endl;
        exit(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;
}
0