結果
問題 | No.856 増える演算 |
ユーザー |
|
提出日時 | 2019-08-01 17:46:37 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
TLE
|
実行時間 | - |
コード長 | 2,544 bytes |
コンパイル時間 | 2,377 ms |
コンパイル使用メモリ | 204,124 KB |
最終ジャッジ日時 | 2025-01-07 10:28:32 |
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 79 TLE * 1 |
ソースコード
#include <bits/stdc++.h>using namespace std;typedef complex<double> cpx;const double PI = acos(-1);vector<cpx> FFT(vector<cpx> P, bool inv = 0) {assert(__builtin_popcount(P.size()) == 1);int lg = 31 - __builtin_clz(P.size()), n = 1 << lg; // == P.size();for (int j = 1, i = 0; j < n - 1; ++j) {for (int k = n >> 1; k > (i ^= k); k >>= 1);if (j < i) swap(P[i], P[j]);} // bit reverseauto w1 = exp((2 - 4 * inv) * PI / n * cpx(0, 1)); // order is 1<<lgfor (int i = 1; i <= lg; ++i) {auto wn = pow(w1, 1 << (lg - i)); // order is 1<<ifor (int k = 0; k < (1 << lg); k += 1 << i) {cpx base = 1;for (int j = 0; j < (1 << i - 1); ++j, base = base * wn) {auto t = base * P[k + j + (1 << i - 1)];auto u = P[k + j];P[k + j] = u + t;P[k + j + (1 << i - 1)] = u - t;}}}if (inv)for (int i = 0; i < n; ++i) P[i] /= n;return P;}const int M = 1e9 + 7;int ipow(int v, int64_t n, int m = M) {int r = 1;while (n) {if (n & 1) r = 1LL * r * v % m;v = 1LL * v * v % m;n >>= 1;}return r;}int inv(int v, int m = M) {return ipow(v, m - 2, m);}int main() {ios::sync_with_stdio(false);int N;cin >> N;vector<int> A(N);for (int i = 0; i < N; ++i) {cin >> A[i];}pair<int, int> dnr(A[0], A[1]);for (int i = N - 2, minp = A[N - 1]; ~i; --i) {auto log_eval = [](int x, int y) {return log(x + y) + y * log(x);};if (log_eval(A[i], minp) < log_eval(dnr.first, dnr.second)) {dnr = make_pair(A[i], minp);}minp = min(minp, A[i]);}vector<int64_t> pa(N + 1);for (int i = 0; i < N; ++i) {pa[i + 1] = pa[i] + A[i];}int maxa = *max_element(A.begin(), A.end());int maxa2 = maxa * 2 + 1; while (__builtin_popcount(maxa2) != 1) ++maxa2;vector<cpx> c(maxa2);for (int a : A) c[a] += 1.0;vector<cpx> fc = FFT(c);for (int i = 0; i < fc.size(); ++i) {fc[i] *= fc[i];}vector<cpx> c2 = FFT(fc, true);int ans = 1;for (int s = 2; s < maxa * 2 + 1; ++s) {int64_t cs = int(real(c2[s]) + 0.5) / 2;if (~s & 1) cs -= int(real(c[s / 2]) + 0.5) / 2;ans = 1LL * ans * ipow(s, cs) % M;}for (int i = 0; i + 1 < N; ++i) {ans = 1LL * ans * ipow(A[i], pa[N] - pa[i + 1]) % M;}int idnr = inv(1LL * (dnr.first + dnr.second) % M * ipow(dnr.first, dnr.second) % M);ans = 1LL * ans * idnr % M;cout << ans << endl;return 0;}