#1198980 (C++17) No.3799 1 + 6 = 7 (Another version)

• 問題: No.3799 1 + 6 = 7 (Another version)
• ユーザー: GOTKAKO
• 提出日時: 2026-11-13 23:19:17
• 言語: C++17 (gcc 15.3.0 + boost 1.92.0 + ACL)
• 結果: AC
• 実行時間: 2 ms / 2,000 ms
• コード長: 4,662 bytes
• コンパイル時間: 1,550 ms
• コンパイル使用メモリ: 228,860 KB
• 実行使用メモリ: 6,400 KB
• 最終ジャッジ日時: 2026-11-13 23:19:22
• 合計ジャッジ時間: 4,336 ms
• ジャッジサーバーID （参考情報）: judge3_0 / judge2_0

ファイルパターン	結果
sample	AC * 3
other	AC * 36

ソースコード

#include <bits/stdc++.h>
using namespace std;

//入力が必ず-mod<=a<modの時.
template<const int mod> //mod<2^30.
struct modint{ //mod変更が不可能.
    public:
    long long v;
    static void setmod(int m){} //飾り.
    static constexpr long long getmod(){return mod;}
    modint():v(0){}
    template<typename T>
    modint(T a):v(a){if(v < 0) v += mod;}
    long long val()const{return v;}
 
    modint &operator=(const modint &b) = default;
    modint &operator+()const{return (*this);}
    modint operator-()const{return modint(0)-(*this);}
    modint operator+(const modint b)const{return modint(v)+=b;}
    modint operator-(const modint b)const{return modint(v)-=b;}
    modint operator*(const modint b)const{return modint(v)*=b;}
    modint operator/(const modint b)const{return modint(v)/=b;}
    modint &operator+=(const modint b){
        v += b.v; if(v >= mod) v -= mod;
        return *this;
    }
    modint &operator-=(const modint b){
        v -= b.v; if(v < 0) v += mod; 
        return *this;
    }   
    modint &operator*=(const modint b){v = v*b.v%mod; return *this;}
    modint &operator/=(modint b){ //b!=0 mod素数が必須.
        assert(b.v != 0);
        (*this) *= b.pow(mod-2);
        return *this;
    }
    modint pow(long long n)const{
        modint ret = 1,p = v;
        if(n < 0) p = p.inv(),n = -n;
        while(n){
            if(n&1) ret *= p;
            p *= p; n >>= 1;
        }
        return ret;
    }
    modint inv()const{return pow(mod-2);} //素数mod必須.
 
    modint &operator++(){*this += 1; return *this;}
    modint &operator--(){*this -= 1; return *this;}
    modint operator++(int){modint ret = *this; *this += 1; return ret;}
    modint operator--(int){modint ret = *this; *this -= 1; return ret;}
    friend bool operator==(const modint a,const modint b){return a.v==b.v;}
    friend bool operator!=(const modint a,const modint b){return a.v!=b.v;}
    friend bool operator<(const modint a,const modint b){return a.v<b.v;}
    friend bool operator<=(const modint a,const modint b){return a.v<=b.v;}
    friend bool operator>=(const modint a,const modint b){return a.v>=b.v;}
    friend bool operator>(const modint a,const modint b){return a.v>b.v;}
    friend ostream &operator<<(ostream &os,const modint a){return os<<a.v;}
    friend istream &operator>>(istream &is,modint &a){ //入力はmodをとってくれる.
        long long x; is >> x; x %= mod;
        a = modint(x); return is;
    }
};
using mint = modint<998244353>; const long long mod = 998244353;

mint solve(long long N,long long M){
    vector<long long> A(N);
    mint ret = 0;
    vector<mint> C(N);
    auto dfs = [&](auto dfs,int pos,long long sum) -> void {
        if(pos == N){
            if(sum%2) return;
            long long now = 0;
            for(int i=0; i<N; i++){
                now += A.at(i);
                if(now+now > sum) break;
                if(now+now == sum){C.at(i)++,ret++; break;}
            }
            return;
        }
        for(int i=1; i<=M; i++){
            A.at(pos) = i;
            dfs(dfs,pos+1,sum+i); 
        }
    };
    dfs(dfs,0,0);

    //for(auto c : C) cout << c << " "; cout << endl;
    return ret;
}

template<typename T>
vector<vector<T>> multima(vector<vector<T>> &A,vector<vector<T>> &B){
    assert(A.at(0).size() == B.size() && B.size());
    int H = A.size(),W = B.at(0).size();
    vector<vector<T>> ret(H,vector<T>(W));
    for(int i=0; i<H; i++) for(int k=0; k<W; k++){
        for(int l=0; l<A.at(i).size(); l++){
            ret.at(i).at(k) += A.at(i).at(l)*B.at(l).at(k);
        }
    }
    return ret; //計算できる行列のみ.
}
template<typename T>
vector<vector<T>> Powmat(vector<vector<T>> A,vector<vector<T>> B,long long N){ //return A^N*B ABは行列;
    //n*n行列A n*1行列Bと想定 O(n^3*logN).
    assert(A.size() && A.size() == A.at(0).size() && N >= 0);
    while(N){
        if(N&1) B = multima(A,B);
        A = multima(A,A); N >>= 1;
    }
    return B;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    long long N,M; cin >> N >> M;
    if(N == 1) cout << "0\n";
    else if(N == 2) cout << M%mod << "\n";
    else{
        mint v = 1;
        M %= mod;
        v *= mint(M);
        v *= mint(M)-1;
        vector<vector<mint>> A = {
            {M*M%mod,1,-1},
            {0,1,0},
            {0,0,1}
        },B = {{v},{M*M%mod},{M}};

        long long n = (N-3)/2;
        while(n){
            if(n&1) B = multima(A,B);
            A = multima(A,A),n >>= 1;
        }
        if((N-3)%2) B.at(0).at(0) = B.at(0).at(0)*M+M;
        cout << B.at(0).at(0) << endl;
    }
}