結果

問題 No.3227 Matrix Query
ユーザー hint908
提出日時 2025-08-08 21:42:55
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,835 ms / 8,000 ms
コード長 8,247 bytes
コンパイル時間 3,379 ms
コンパイル使用メモリ 307,728 KB
実行使用メモリ 11,136 KB
最終ジャッジ日時 2025-08-08 21:43:40
合計ジャッジ時間 32,201 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 * 1
other AC * 28
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("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));
    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;
}

template <typename T, T (*OP)(T, T), T (*E)()>
struct segtree { 
    vector<T> seg;
    int seg_size, _n;//葉の数
    
    segtree(int N) : seg(), seg_size(), _n(N) {
        seg_size = 1;
        while(seg_size < N) seg_size <<= 1;
        seg.resize(seg_size<<1, E());
    }
    segtree(const vector<T> a) : seg(), seg_size(), _n(a.size()) {
        int N = a.size();
        seg_size = 1;
        while(seg_size < N) seg_size <<= 1;
        seg.resize(seg_size<<1, E());
        for(int i=0; i<N; i++) seg[i + seg_size] = a[i];
        for(int i = seg_size-1; i > 0; i --){
            seg[i] = OP(seg[(i<<1)], seg[(i<<1)|1]);
        }
    }
    
    void set(int idx, T x){
        idx += seg_size;
        seg[idx] = x;
        update(idx);
    }
    T get(int idx){
        idx += seg_size;
        return seg[idx];
    }
    
    void apply(int idx, T x){
        idx += seg_size;
        seg[idx] = OP(seg[idx], x);
        update(idx);
    }

    void update(int idx){
        while(idx > 1){
            idx >>= 1;
            seg[idx] = OP(seg[idx<<1], seg[(idx<<1)|1]);
        }
    }
    
    
    T prod(int left, int right) {
        if(left < 0) left = 0;
        if(right > _n) right = _n;
        if(left > right) return E();

        left += seg_size;
        right += seg_size;

        T retl= E(), retr = E();
        {
            int L = left, R = right;
            while(L < R){
                if(L&1){
                    retl = OP(retl, seg[L]);
                    L++;
                }
                if(R&1){
                    R--;
                    retr = OP(seg[R], retr);
                }
                L >>= 1;
                R >>= 1;
            }
        }
        return OP(retl, retr);
    }
    T all_prod(){ return seg[1]; }

    template<class F>
    int max_right(int left, F f) {
        if(left >= _n) return _n;
        if(left < 0) left = 0;
        T s = E();
        left += seg_size;
        do{
            while(left % 2 == 0) left >>= 1;
            if(!f(OP(s, seg[left]))){
                while(left < seg_size){
                    left <<= 1;
                    if(f(OP(s, seg[left]))){
                        s = OP(s, seg[left]);
                        left++;
                    }
                }
                return left - seg_size;
            }
            s = OP(s, seg[left]);
            left++;
        }while((left & (left-1)));
        return _n;
    };

    template<class F>
    int min_left(int right, F f) {
        if(right <= 0) return 0;
        if(right >= _n) right = _n;
        T s = E();
        right += seg_size;
        do{
            while(right % 2 == 0) right >>= 1;
            if(right != 1) right --;
            if(!f(OP(seg[right], s))){
                while(right < seg_size){
                    right <<= 1;
                    right |= 1;
                    if(f(OP(seg[right], s))){
                        s = OP(seg[right], s);
                        right ^= 1;
                    }
                }
                return right - seg_size;
            }
            s = OP(seg[right], s);
        }while((right & (right-1)));
        return 0;
    };
};

/*
ll e(){return -INF;}
ll op(ll L, ll R){return max(L, R);}
*/


ll K;
VV<ll> op(VV<ll> a, VV<ll> b){return matmul(a,b,K);}
VV<ll> e(){return {{1,0},{0,1}};}

void solve(){
	ll n;
	cin >> K >> n;
	VVV<ll> v;
	rep(i,n){
		VV<ll> w(2, V<ll>(2));
		rep(i,2) rep(j,2) cin >> w[i][j];
		v.eb(w);
	}
	segtree<VV<ll>,op,e> seg(v);
	ll q;
	cin >> q;
	rep(i,q){
		ll x,l,r;
		cin >> x >> l >> r;
		l--;
		x--;
		VV<ll> w(2, V<ll>(2));
		rep(i, 2) rep(j, 2) cin >> w[i][j];
		seg.set(x, w);
		auto ans = seg.prod(l, r);
		rep(i, 2) rep(j, 2){
			(ans[i][j] += K) %= K;
			cout << ans[i][j] << " \n"[j];
		}
	}
    
}

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