結果

問題 No.2792 Security Cameras on Young Diagram
ユーザー k1suxu
提出日時 2024-06-22 14:48:27
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 34 ms / 2,000 ms
コード長 8,087 bytes
コンパイル時間 4,903 ms
コンパイル使用メモリ 275,108 KB
実行使用メモリ 7,224 KB
最終ジャッジ日時 2024-06-24 18:48:48
合計ジャッジ時間 6,126 ms
ジャッジサーバーID
(参考情報)
judge5 / judge2
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 * 21
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

#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
#define FOR(n) for(int i = 0; i < (int)n; i++)
#define repi(i,a,b) for(int i = (int)a; i < (int)b; i++)
#define all(x) x.begin(),x.end()
//#define mp make_pair
#define vi vector<int>
#define vvi vector<vi>
#define vvvi vector<vvi>
#define vvvvi vector<vvvi>
#define pii pair<int,int>
#define vpii vector<pair<int,int>>
template<typename T>
bool chmax(T &a, const T b) {if(a<b) {a=b; return true;} else {return false;}}
template<typename T>
bool chmin(T &a, const T b) {if(a>b) {a=b; return true;} else {return false;}}
using ll = long long;
using ld = long double;
using ull = unsigned long long;
const ll INF = numeric_limits<long long>::max() / 2;
const ld pi = 3.1415926535897932384626433832795028;
const ll mod = 998244353;
int dx[] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};
#define int long long
template<long long MOD>
struct Modular_Int {
using Mint_Type = Modular_Int<MOD>;
long long x;
Modular_Int() = default;
Modular_Int(long long x_) : x(x_ >= 0? x_%MOD : (MOD-(-x_)%MOD)%MOD) {}
long long val() const {
return (x%MOD+MOD)%MOD;
}
static long long get_mod() {
return MOD;
}
Mint_Type& operator^=(long long d) {
Mint_Type ret(1);
long long nx = x;
while(d) {
if(d&1) ret *= nx;
(nx *= nx) %= MOD;
d >>= 1;
}
*this = ret;
return *this;
}
Mint_Type operator^(long long d) const {return Mint_Type(*this) ^= d;}
Mint_Type pow(long long d) const {return Mint_Type(*this) ^= d;}
//use this basically
Mint_Type inv() const {
return Mint_Type(*this) ^ (MOD-2);
}
//only if the module number is not prime
//Don't use. This is broken.
// Mint_Type inv() const {
// long long a = (x%MOD+MOD)%MOD, b = MOD, u = 1, v = 0;
// while(b) {
// long long t = a/b;
// a -= t*b, swap(a, b);
// u -= t*v, swap(u, v);
// }
// return Mint_Type(u);
// }
Mint_Type& operator+=(const Mint_Type other) {
if((x += other.x) >= MOD) x -= MOD;
return *this;
}
Mint_Type& operator-=(const Mint_Type other) {
if((x -= other.x) < 0) x += MOD;
return *this;
}
Mint_Type& operator*=(const Mint_Type other) {
long long z = x;
z *= other.x;
z %= MOD;
x = z;
if(x < 0) x += MOD;
return *this;
}
Mint_Type& operator/=(const Mint_Type other) {
return *this = *this * other.inv();
}
Mint_Type& operator++() {
x++;
if (x == MOD) x = 0;
return *this;
}
Mint_Type& operator--() {
if (x == 0) x = MOD;
x--;
return *this;
}
Mint_Type operator+(const Mint_Type other) const {return Mint_Type(*this) += other;}
Mint_Type operator-(const Mint_Type other) const {return Mint_Type(*this) -= other;}
Mint_Type operator*(const Mint_Type other) const {return Mint_Type(*this) *= other;}
Mint_Type operator/(const Mint_Type other) const {return Mint_Type(*this) /= other;}
Mint_Type& operator+=(const long long other) {Mint_Type other_(other); *this += other_; return *this;}
Mint_Type& operator-=(const long long other) {Mint_Type other_(other); *this -= other_; return *this;}
Mint_Type& operator*=(const long long other) {Mint_Type other_(other); *this *= other_; return *this;}
Mint_Type& operator/=(const long long other) {Mint_Type other_(other); *this /= other_; return *this;}
Mint_Type operator+(const long long other) const {return Mint_Type(*this) += other;}
Mint_Type operator-(const long long other) const {return Mint_Type(*this) -= other;}
Mint_Type operator*(const long long other) const {return Mint_Type(*this) *= other;}
Mint_Type operator/(const long long other) const {return Mint_Type(*this) /= other;}
bool operator==(const Mint_Type other) const {return (*this).val() == other.val();}
bool operator!=(const Mint_Type other) const {return (*this).val() != other.val();}
bool operator==(const long long other) const {return (*this).val() == other;}
bool operator!=(const long long other) const {return (*this).val() != other;}
Mint_Type operator-() const {return Mint_Type(0LL)-Mint_Type(*this);}
//-1: sqrt
//
long long get_sqrt() const {
long long a = val(), p = get_mod();
if(a == 0) return 0;
if(p == 2) return a;
if(Mint_Type(a).pow((p - 1) >> 1).val() != 1) return -1;
long long b = 1;
while(Mint_Type(b).pow((p - 1) >> 1).val() == 1) ++b;
long long e = 0, m = p - 1;
while(m % 2 == 0) m >>= 1, ++e;
long long x = Mint_Type(a).pow((m - 1) >> 1).val();
long long y = a * (x * x % p) % p;
(x *= a) %= p;
long long z = Mint_Type(b).pow(m).val();
while(y != 1) {
long long j = 0, t = y;
while(t != 1) {
j += 1;
(t *= t) %= p;
}
z = Mint_Type(z).pow((long long)1 << (e - j - 1)).val();
(x *= z) %= p;
(z *= z) %= p;
(y *= z) %= p;
e = j;
}
return x;
}
template <typename T>
friend Mint_Type operator+(T t, const Mint_Type& o) {
return o + t;
}
template <typename T>
friend Mint_Type operator-(T t, const Mint_Type& o) {
return -o + t;
}
template <typename T>
friend Mint_Type operator*(T t, const Mint_Type& o) {
return o * t;
}
template <typename T>
friend Mint_Type operator/(T t, const Mint_Type& o) {
return o.inv() * t;
}
};
// TODO: SELECT MOD_VAL
// const long long MOD_VAL = 1e9+7;
const long long MOD_VAL = 998244353;
using mint = Modular_Int<MOD_VAL>;
istream& operator>>(istream& is, mint& x) {
long long X;
is >> X;
x = X;
return is;
}
ostream& operator<<(ostream& os, mint& x) {
os << x.val();
return os;
}
// 1e9 + 7mod使CFCCAtCoder
vector<mint> fact = {1}, fact_inv = {1};
void factor_init(long long n) {
++n;
fact.resize(n, 0);
fact_inv.resize(n, 0);
fact[0] = 1;
repi(i, 1, n) fact[i] = (fact[i - 1] * i);
fact_inv[n-1] = fact[n-1].inv();
for(int i = n-1; i > 0; --i) fact_inv[i-1] = fact_inv[i] * i;
}
mint P(long long n, long long k) {
if(n<k) {
cerr << "\nAssertion Failed!!\n";
cerr << "Expression N >= K, ";
cerr << "where n=" << n << ",k=" << k << "\n\n";
return 0;
}
while(n > fact.size()-1) {
fact.push_back(fact.back() * fact.size());
fact_inv.push_back(fact.back().inv());
}
return fact[n] * fact_inv[n-k];
}
mint C(long long n, long long k) {
if(n<k) {
cerr << "\nAssertion Failed!!\n";
cerr << "Expression N >= K, ";
cerr << "where n=" << n << ",k=" << k << "\n\n";
return 0;
}
while(n > fact.size()-1) {
fact.push_back(fact.back() * fact.size());
fact_inv.push_back(fact.back().inv());
}
return fact[n]*fact_inv[n-k]*fact_inv[k];
}
mint H(long long n, long long k) {
assert(n>=1);
return C(n+k-1, k);
}
mint Cat(long long n) {
return C(2*n, n)-C(2*n, n-1);
}
void solve() {
const int MX = 2e5;
int n;
cin >> n;
vi a(n+1);
FOR(n) cin >> a[i];
a[n] = 0;
mint ans = 0;
FOR(n) repi(j, a[i+1]+1, a[i]+1) ans += C(i+j-1, i);
vi tot(MX+1);
FOR(n) tot[a[i]]++;
for (int i = MX; i > 0; i--) tot[i-1] += tot[i];
FOR(MX) repi(j, tot[i+1]+1, tot[i]+1) ans += C(i+j-2, i-1);
cout << ans << endl;
}
signed main() {
cin.tie(nullptr);
ios::sync_with_stdio(false);
solve();
return 0;
}
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