#1007502 (C++17) No.2872 Depth of the Parentheses

• 問題: No.2872 Depth of the Parentheses
• ユーザー: GOTKAKO
• 提出日時: 2024-09-06 21:45:13
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 647 ms / 2,000 ms
• コード長: 2,524 bytes
• コンパイル時間: 1,831 ms
• コンパイル使用メモリ: 195,644 KB
• 最終ジャッジ日時: 2025-02-24 04:10:51
• ジャッジサーバーID （参考情報）: judge5 / judge1

ファイルパターン	結果
sample	AC * 3
other	AC * 20 WA * 5

ソースコード

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#include <bits/stdc++.h>
using namespace std;
long long mod = 998244353;
//入力が必ず-mod<a<modの時.
struct mint{
    long long v = 0;
    mint(){} mint(int a){v = a<0?a+mod:a;} mint(long long a){v = a<0?a+mod:a;}
    mint(unsigned long long a){v = a;}
    long long val(){return v;}
    void modu(){v %= mod;}
 
    mint repeat2mint(const long long a,long long b){
        mint ret = 1,p = a;
        while(b){
            if(b&1) ret *= p;
            p *= p; b >>= 1;
        }
        return ret;
    };
 
    mint &operator=(const mint &b) = default;
    mint operator-() const {return mint(0)-(*this);}
    mint operator+(const mint b){return mint(v)+=b;}
    mint operator-(const mint b){return mint(v)-=b;}
    mint operator*(const mint b){return mint(v)*=b;}
    mint operator/(const mint b){return mint(v)/=b;}
 
    mint operator+=(const mint b){
        v += b.v; if(v >= mod) v -= mod;
        return *this;
    }
    mint operator-=(const mint b){
        v -= b.v; if(v < 0) v += mod; 
        return *this;
    }   
    mint operator*=(const mint b){v = v*b.v%mod; return *this;}
    mint operator/=(mint b){
        if(b == 0) assert(false);
        int left = mod-2;
        while(left){if(left&1) *this *= b; b *= b; left >>= 1;}
        return *this;
    }
 
    mint operator++(int){*this += 1; return *this;}
    mint operator--(int){*this -= 1; return *this;}
    bool operator==(const mint b){return v == b.v;}
    bool operator!=(const mint b){return v != b.v;}
    bool operator>(const mint b){return v > b.v;}
    bool operator>=(const mint b){return v >= b.v;}
    bool operator<(const mint b){return v < b.v;}
    bool operator<=(const mint b){return v <= b.v;}
 
    mint pow(const long long x){return repeat2mint(v,x);}
    mint inv(){return mint(1)/v;}
};
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int x,K; cin >> x >> K;
    if(K > 10){cout << "how?" << endl; return 0;}
    int n2 = (1<<(K*2));
    mint answer = 0,div100 = mint(1)/100;
    for(int i=0; i<n2; i++){
        string now = "";
        for(int k=0; k<2*K; k++){
            if(i&(1<<k)) now += ')';
            else now += '(';
        }
        mint p = 1;
        int nowd = 0,maxd = 0,ok = 1;
        for(auto &c : now){
            if(c == '(') nowd++,maxd = max(maxd,nowd),p *= div100*x;
            else nowd--,p *= div100*(100-x);
            if(nowd < 0) ok = 0;
        }
        if(nowd != 0) ok = 0;
        if(ok) answer += p*maxd;
    }
    cout << answer.v << "\n";
}
 
 
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