結果

問題 No.3182 recurrence relation’s intersection sum
ユーザー hint908
提出日時 2025-06-13 22:12:10
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
(最新)
AC  
(最初)
実行時間 -
コード長 5,210 bytes
コンパイル時間 3,679 ms
コンパイル使用メモリ 294,312 KB
実行使用メモリ 13,824 KB
最終ジャッジ日時 2025-06-23 00:55:48
合計ジャッジ時間 33,706 ms
ジャッジサーバーID
（参考情報） 		judge2 / 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 * 39 TLE * 1
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ソースコード

diff # 

// #pragma GCC target("avx2")
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")


#include<bits/stdc++.h>
using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template<class T> using V = vector<T>;
template<class T> using VV = V<V<T>>;
template<class T> using VVV = V<VV<T>>;
template<class T> using VVVV = VV<VV<T>>;
#define rep(i,n) for(ll i=0ll;(i)<(n);(i)++)
#define REP(i,a,n) for(ll i=(a);(i)<(n);(i)++)
#define rrep(i,n) for(ll i=(n)-1;(i)>=(0ll);(i)--)
#define RREP(i,a,n) for(ll i=(n)-1;(i)>=(a);(i)--)
const long long INF = (1LL << 60);
const long long mod99 = 998244353;
const long long mod107 = 1000000007;
const long long mod = mod99;
#define eb emplace_back
#define be(v) (v).begin(),(v).end()
#define all(v) (v).begin(),(v).end()
#define foa(i,v) for(auto& (i) : (v))
#define UQ(v) sort(be(v)), (v).erase(unique(be(v)), (v).end())
#define UQ2(v,cmp) sort(be(v)), (v).erase(unique(be(v),cmp), (v).end())
#define UQ3(v,cmp) sort(be(v),cmp), (v).erase(unique(be(v)), (v).end())
#define UQ4(v,cmp,cmp2) sort(be(v), cmp), (v).erase(unique(be(v),cmp2), (v).end())
#define LB(x,v) (lower_bound(be(v),(x))-(v).begin())
#define LB2(x,v,cmp) (lower_bound(be(v),(x),(cmp))-(v).begin())
#define UB(x,v) (upper_bound(be(v),(x))-(v).begin())
#define UB2(x,v,cmp) (upper_bound(be(v),(x),(cmp))-(v).begin())
#define dout()  cout << fixed << setprecision(20)
#define randinit() srand((unsigned)time(NULL))

template<class T, class U> bool chmin(T& t, const U& u) { if (t > u){ t = u; return 1;} return 0; }
template<class T, class U> bool chmax(T& t, const U& u) { if (t < u){ t = u; return 1;} return 0; }


ll Rnd(ll L=0, ll R=mod99){return rand()%(R-L)+L;}


VV<ll> matmul(VV<ll> v, VV<ll> w, ll p=(1ll<<60)){
    ll n1 = v.size();
    ll n2 = w.size();
    ll n3 = w[0].size();
    VV<ll> ret(n1, V<ll>(n3, 0));
    rep(i, n1) rep(j,n2) rep(k,n3) (ret[i][k] += v[i][j]*w[j][k]) %= p;
    return ret;
}

VV<ll> matpow(VV<ll> v, ll k, ll p){
    if(k == 1) return v;
    ll n = v.size();
    VV<ll> ret(n, V<ll>(n, 0));
    if(k==-1) return ret;
    rep(i, n) ret[i][i] = 1;
    if(k == 0) return ret;
    
    VV<ll> w = matpow(v, k/2, p);
    w = matmul(w, w, p);
    if(k%2) w = matmul(w, v, p);
    
    return w;
    
}


struct Combination{
    vector<long long> fac, inv, finv;
    long long MOD;
    Combination(long long N = 200100, long long p = 998244353) : fac(N, 1), inv(N, 1), finv(N, 1), MOD(p){
        for(long long i = 2; i < N; i++){
            fac[i] = fac[i-1] * i % MOD;
            inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
            finv[i] = finv[i-1] * inv[i] % MOD;
        }
    }
    long long com(long long n, long long k){
        if(n < k) return 0;
        if(n < 0 || k < 0) return 0;
        return fac[n] * finv[k] % MOD * finv[n-k] % MOD;
    }
    
    long long per(long long n, long long k){
        if(n < k) return 0;
        if(n < 0 || k < 0) return 0;
        return fac[n] * finv[n-k] % MOD;
    }
};

long long modpow(long long n, long long k, long long p = mod){
    long long a = n % p;
    long long ans = 1;
    while(k != 0) {
        if(k & 1) ans = ans * a % p;
        k /= 2;
        a = a * a % p;
    }
    
    return ans;  
}

// n^(-1) ≡ b (mod p) となる b を求める
long long modinv(long long n, long long p = mod) { 
//    if(n == 1) return 1;
//    return p - modinv(p % n) * (p / n) % p;
    return modpow(n, p - 2, p);
}

// n^k ≡ b (mod p) となる最小の k を求める
long long modlog(long long n, long long b, long long p = mod){
  
    long long sqrt_p = sqrt(p);
    unordered_map<long long , long long> n_pow;
    long long memo = 1;
    
    for(long long i = 0; i < sqrt_p; i ++){
        if(!n_pow.count(memo)) n_pow[memo] = i;
        memo = memo * n % p; 
    }
    
    memo = modinv(memo, p);
    long long ans = 0;
    while(!n_pow.count(b)){
        if(ans >= p) return -1;
        ans += sqrt_p;
        b = b * memo % p;
    }
  
    ans += n_pow[b];
    return ans % (p - 1);

}

// ax + by = gcd(a, b) を満たす (x, y) が格納される
long long ext_gcd(long long a, long long b, long long &x, long long &y){
    if(b == 0){
        x = 1;
        y = 0;
        return a;
    }
    long long d = ext_gcd(b, a%b, y, x);
    y -= a/b*x;
    return d;
}

void solve(){
    ll k,l,r;
    cin >> k >> l >> r;
    Combination com(k+10);
    
    VV<ll> vv(k+4, V<ll>(k+4, 0));
    rep(i, k+1){
        rep(j,i+1) vv[i][j] = com.com(i, j);
    }
    vv[k+1][k+1] = k;
    vv[k+2][k+2] = k;
    vv[k+2][k] = 1;
    vv[k+2][k+1] = 1;
    vv[k+3] = vv[k+2];
    vv[k+3][k+3] = 1;
    
    auto f = matpow(vv, r, mod);
    auto g = matpow(vv, l-1, mod);
    // VV<ll> w(k+4, 1);
    ll ans = f[k+3][0] - g[k+3][0];
    // ans += f[k+3][k+2] - g[k+3][k+2];
    // ans += f[k+3][k+3] - g[k+3][k+3];;
    REP(i, k+1, k+4) ans += f[k+3][i] - g[k+3][i];
    
    // rep(i, k+4) rep(j, k+4) cout << vv[i][j] << " \n"[j==k+3];
    ans %= mod;
    (ans += mod) %= mod;
    cout << ans << endl;
    
}





int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int t=1;
    // cin >> t;
    rep(i,t) solve();
}
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