結果

問題 No.1084 積の積
ユーザー outline
提出日時 2020-12-11 18:34:35
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,780 bytes
コンパイル時間 1,583 ms
コンパイル使用メモリ 131,504 KB
最終ジャッジ日時 2025-01-16 22:19:29
ジャッジサーバーID
(参考情報)
judge3 / judge4
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 * 5
other AC * 20 WA * 7
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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <stack>
#include <tuple>
#include <deque>
#include <array>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <chrono>
#include <random>
#include <limits>
#include <iterator>
#include <functional>
#include <sstream>
#include <fstream>
#include <complex>
#include <cstring>
#include <unordered_map>
using namespace std;
using ll = long long;
using P = pair<int, int>;
constexpr int INF = 1001001001;
constexpr int mod = 1000000007;
// constexpr int mod = 998244353;
template<class T>
inline bool chmax(T& x, T y){
if(x < y){
x = y;
return true;
}
return false;
}
template<class T>
inline bool chmin(T& x, T y){
if(x > y){
x = y;
return true;
}
return false;
}
struct mint {
int x;
mint() : x(0) {}
mint(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
mint& operator+=(const mint& p){
if((x += p.x) >= mod) x -= mod;
return *this;
}
mint& operator-=(const mint& p){
if((x -= p.x) < 0) x += mod;
return *this;
}
mint& operator*=(const mint& p){
x = (int)(1LL * x * p.x % mod);
return *this;
}
mint& operator/=(const mint& p){
*this *= p.inverse();
return *this;
}
mint operator-() const { return mint(-x); }
mint operator+(const mint& p) const { return mint(*this) += p; }
mint operator-(const mint& p) const { return mint(*this) -= p; }
mint operator*(const mint& p) const { return mint(*this) *= p; }
mint operator/(const mint& p) const { return mint(*this) /= p; }
bool operator==(const mint& p) const { return x == p.x; }
bool operator!=(const mint& p) const { return x != p.x; }
mint pow(int64_t n) const {
mint res = 1, mul = x;
while(n > 0){
if(n & 1) res *= mul;
mul *= mul;
n >>= 1;
}
return res;
}
mint inverse() const { return pow(mod - 2); }
friend ostream& operator<<(ostream& os, const mint& p){
return os << p.x;
}
friend istream& operator>>(istream& is, mint& p){
int64_t val;
is >> val;
p = mint(val);
return is;
}
};
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int N;
cin >> N;
vector<int> A(N);
for(int i = 0; i < N; ++i) cin >> A[i];
// [L, R) i ([i,i])
// (i-L+1)*(R-i) -1 = -i^2 + (L+R-1)*i + (-LR+R-1)
vector<ll> coeff_2(N + 1), coeff_1(N + 1), coeff_0(N + 1);
constexpr int bound = 1e+9;
ll prod = A[0];
int l = 0, r = 1;
while(r < N){
if(prod * A[r] >= bound){
coeff_2[l] += -1, coeff_2[r] -= -1;
coeff_1[l] += l + r - 1, coeff_1[r] -= l + r - 1;
coeff_0[l] += ll(-l) * r + r - 1, coeff_0[r] -= ll(-l) * r + r - 1;
while(l < r && prod * A[r] >= bound){
prod /= A[l++];
}
}
prod *= A[r++];
}
coeff_2[l] += -1, coeff_2[r] -= -1;
coeff_1[l] += l + r - 1, coeff_1[r] -= l + r - 1;
coeff_0[l] += ll(-l) * r + r - 1, coeff_0[r] -= ll(-l) * r + r - 1;
for(int i = 0; i < N; ++i){
coeff_0[i + 1] += coeff_0[i];
coeff_1[i + 1] += coeff_1[i];
coeff_2[i + 1] += coeff_2[i];
}
vector<ll> W(N);
for(int i = 0; i < N; ++i){
W[i] = 1 + coeff_2[i] * i * i + coeff_1[i] * i + coeff_0[i];
}
mint ans = 1;
for(int i = 0; i < N; ++i){
ans *= mint(A[i]).pow(W[i]);
}
cout << ans << '\n';
return 0;
}
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