結果

問題 No.856 増える演算
ユーザー fumiphys
提出日時 2019-08-10 16:37:25
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 7,036 bytes
コンパイル時間 2,157 ms
コンパイル使用メモリ 182,804 KB
実行使用メモリ 17,936 KB
最終ジャッジ日時 2024-07-19 17:09:21
合計ジャッジ時間 16,109 ms
ジャッジサーバーID
(参考情報)
judge1 / judge3
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

// includes
#include <bits/stdc++.h>
// macros
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define vrep(v, i) for(int i = 0; i < (v).size(); i++)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define FI first
#define SE second
#define bit(n) (1LL<<(n))
#define INT(n) int n; cin >> n;
#define LL(n) ll n; cin >> n;
#define DOUBLE(n) double n; cin >> n;
using namespace std;
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size
    ())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr =
    itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr;
    auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto
    titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os <<
    *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os <<
    itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end();
    ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
// types
using ll = long long int;
using P = pair<int, int>;
using Pli = pair<ll, int>;
using Pil = pair<int, ll>;
using Pll = pair<ll, ll>;
using Pdd = pair<double, double>;
using cd = complex<double>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// solve
template <typename T>
T power(T a, T n, T mod) {
n = n % (mod - 1);
T res = 1;
T tmp = n;
T curr = a;
while(tmp){
if(tmp % 2 == 1){
res = (T)(res * curr % mod);
}
curr = (T)(curr * curr % mod);
tmp >>= 1;
}
return res;
}
template<typename T>
T extgcd(T a, T b, T &x, T &y){
T d = a;
if(b != 0){
d = extgcd(b, a % b, y, x);
y -= (a / b) * x;
}else{
x = 1, y = 0;
}
return d;
}
template <typename T>
T modinv(T a, T m){
long long x = 0, y = 0;
extgcd<long long>(a, m, x, y);
x %= m;
if(x < 0)x += m;
return x;
}
template <int MOD, int g>
struct NTT{
int get_mod(){
return MOD;
}
void _ntt(vector<long long> &f, bool inv=false){
int n = f.size(), mask = n - 1;
int h = power<long long>(g, (MOD - 1) / n, MOD);
if(inv)h = modinv(h, MOD);
vector<long long> tmp(n);
for(int i = n >> 1; i >= 1; i >>= 1){
long long zeta = power<long long>(h, i, MOD);
long long w = 1;
for(int j = 0; j < n; j += i){
for(int k = 0; k < i; k++){
tmp[j+k] = (f[((j<<1)&mask)+k] + w * f[(((j<<1)+i)&mask)+k] % MOD) % MOD;
}
w = w * zeta % MOD;
}
swap(f, tmp);
}
}
void ntt(vector<long long> &f){
_ntt(f, false);
}
void intt(vector<long long> &f){
_ntt(f, true);
int n = f.size();
int ni = modinv(n, MOD);
for(int i = 0; i < n; i++)f[i] = f[i] * ni % MOD;
}
vector<long long> convolution(vector<long long> f, vector<long long> h){
int n = 1;
while(n < int(f.size() + h.size()))n *= 2;
f.resize(n, 0); h.resize(n, 0);
ntt(f);
ntt(h);
for(int i = 0; i < n; i++)f[i] = f[i] * h[i] % MOD;
intt(f);
return f;
}
};
using NTT1 = NTT<167772161, 3>;
using NTT2 = NTT<469762049, 3>;
using NTT3 = NTT<1224736769, 3>;
template <typename T>
long long garner(vector<T> b, vector<T> m, T MOD){
m.emplace_back(MOD);
vector<long long> coef(m.size(), 1);
vector<long long> consts(m.size(), 0);
for(int i = 0; i < b.size(); i++){
long long t = ((b[i] - consts[i]) % m[i]) * modinv<long long>(coef[i], m[i]) % m[i];
for(int j = i + 1; j < m.size(); j++){
consts[j] = (consts[j] + t * coef[j] % m[j]) % m[j];
coef[j] = coef[j] * m[i] % m[j];
}
}
return consts.back();
}
vector<long long> arbitrary_mod_convolution(vector<long long> f, vector<long long> g, int MOD){
for(size_t i = 0; i < f.size(); i++)f[i] %= MOD;
for(size_t i = 0; i < g.size(); i++)g[i] %= MOD;
NTT1 ntt1;
NTT2 ntt2;
NTT3 ntt3;
auto x1 = ntt1.convolution(f, g);
auto x2 = ntt2.convolution(f, g);
auto x3 = ntt3.convolution(f, g);
vector<long long> res(x1.size());
vector<long long> b(3), m(3);
m[0] = ntt1.get_mod();
m[1] = ntt2.get_mod();
m[2] = ntt3.get_mod();
for(size_t i = 0; i < x1.size(); i++){
b[0] = x1[i];
b[1] = x2[i];
b[2] = x3[i];
res[i] = garner<long long>(b, m, MOD);
}
return res;
}
int main(int argc, char const* argv[])
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout << fixed << setprecision(20);
INT(n); vector<ll> a(n); cin >> a;
ll res = 1;
ll sum = a[n-1];
rrep(i, n - 1){
res = res * power<ll>(a[i], sum, mod) % mod;
sum += a[i];
}
vector<ll> b(2e5+1, 0);
rep(i, n)b[a[i]]++;
NTT1 ntt1;
auto f = ntt1.convolution(b, b);
rep(i, n)f[a[i] * 2]--;
rep(i, sz(f))f[i] /= 2;
/*rep(i, 10)cout << b[i] << "\n "[i + 1 != 10];
rep(i, 10)cout << b[i] << "\n "[i + 1 != 10];
rep(i, 10)cout << f[i] << "\n "[i + 1 != 10];*/
FOR(i, 1, sz(f)){
res = res * power<ll>(i, f[i], mod) % mod;
}
ll am = a[n-1];
double mini = 1e18;
ll prod = 1;
rrep(i, n - 1){
double tmp = log(a[i] + am) + am * log(a[i]);
if(tmp < mini){
mini = tmp;
prod = (am + a[i]) * power<ll>(a[i], am, mod) % mod;
}
am = min(am, a[i]);
}
res = res * modinv<ll>(prod, mod) % mod;
cout << res << endl;
return 0;
}
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