結果

問題 No.856 増える演算
ユーザー 0w1
提出日時 2019-08-01 17:52:54
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 2,552 bytes
コンパイル時間 2,444 ms
コンパイル使用メモリ 203,536 KB
最終ジャッジ日時 2025-01-07 10:28:46
ジャッジサーバーID
(参考情報)
judge2 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 79 WA * 1
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 #
プレゼンテーションモードにする

#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 reverse
auto w1 = exp((2 - 4 * inv) * PI / n * cpx(0, 1)); // order is 1<<lg
for (int i = 1; i <= lg; ++i) {
auto wn = pow(w1, 1 << (lg - i)); // order is 1<<i
for (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 = int64_t(real(c2[s]) + 0.5) / 2;
if (~s & 1) cs -= int64_t(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;
}
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