結果

問題 No.2792 Security Cameras on Young Diagram
ユーザー k1suxuk1suxu
提出日時 2024-06-22 14:48:27
言語 C++23
(gcc 12.3.0 + boost 1.83.0)
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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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 4 ms
5,248 KB
testcase_01 AC 4 ms
5,376 KB
testcase_02 AC 4 ms
5,376 KB
testcase_03 AC 4 ms
5,376 KB
testcase_04 AC 4 ms
5,376 KB
testcase_05 AC 5 ms
5,376 KB
testcase_06 AC 4 ms
5,376 KB
testcase_07 AC 5 ms
5,376 KB
testcase_08 AC 4 ms
5,376 KB
testcase_09 AC 5 ms
5,376 KB
testcase_10 AC 4 ms
5,376 KB
testcase_11 AC 5 ms
5,376 KB
testcase_12 AC 30 ms
6,992 KB
testcase_13 AC 28 ms
6,692 KB
testcase_14 AC 29 ms
6,968 KB
testcase_15 AC 26 ms
6,596 KB
testcase_16 AC 30 ms
7,012 KB
testcase_17 AC 32 ms
7,028 KB
testcase_18 AC 25 ms
6,808 KB
testcase_19 AC 25 ms
6,704 KB
testcase_20 AC 27 ms
6,792 KB
testcase_21 AC 4 ms
5,376 KB
testcase_22 AC 4 ms
5,376 KB
testcase_23 AC 34 ms
7,224 KB
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 #

#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が存在しない
    //複数存在する場合どれを返すかは不明
    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 + 7をmodとして使いたいときに注意!!!!特にCFやCCなどのAtCoder以外

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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