結果

問題 No.426 往復漸化式
ユーザー daddy
提出日時 2022-12-07 21:22:26
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,849 ms / 5,000 ms
コード長 5,215 bytes
コンパイル時間 2,768 ms
コンパイル使用メモリ 197,272 KB
実行使用メモリ 132,352 KB
最終ジャッジ日時 2024-10-14 01:27:11
合計ジャッジ時間 28,011 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 * 1
other AC * 22
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 #
プレゼンテーションモードにする

#pragma GCC optimization ("O3")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pll = pair<ll,ll>;
#define INF (1LL<<61)
#define MOD 1000000007LL 
// #define MOD 998244353LL
#define EPS (1e-10)
#define PR(x) cout << (x) << endl
#define PS(x) cout << (x) << " "
#define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i))
#define FORE(i,v) for(auto (i):v)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((ll)(x).size())
#define REV(x) reverse(ALL((x)))
#define ASC(x) sort(ALL((x)))
#define DESC(x) {ASC((x)); REV((x));}
#define BIT(s,i) (((s)>>(i))&1)
#define pb push_back
#define fi first
#define se second
template<class T> inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;}
template<class T> inline int chmax(T& a, T b) {if(a<b) {a=b; return 1;} return 0;}
class mint {
public:
    ll x;
    mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
    mint operator-() const {return mint(-x);}
    mint& operator+=(const mint& a) {if((x+=a.x)>=MOD) x-=MOD; return *this;}
    mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;}
    mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;}
    mint operator+(const mint& a) const {mint b(*this); return b+=a;}
    mint operator-(const mint& a) const {mint b(*this); return b-=a;}
    mint operator*(const mint& a) const {mint b(*this); return b*=a;}
    mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;}
    mint inv() const {return pow(MOD-2);}
    mint& operator/=(const mint& a) {return *this*=a.inv();}
    mint operator/(const mint& a) const {mint b(*this); return b/=a;}
};
istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;}
ostream &operator<<(ostream& os, const mint& a) {return os<<a.x;}
template <typename X> struct SegmentTree {    
    
    using FX = function<X(X, X)>;
 
    int n;
    FX fx;
    const X ex;
    vector<X> dat;
    
    SegmentTree(int n_, FX fx_, X ex_)
        : n(), fx(fx_), ex(ex_) {
        int x = 1;
        while(n_ > x) x *= 2;
        n = x;
        dat = vector<X>(n*2, ex);
    }
    void set(int i, X x) {
        dat[i+n-1] = x;
    }
    
    void build() {
        for (int k = n-2; k>=0; k--) dat[k] = fx(dat[k*2+1], dat[k*2+2]);
    }
    
    void update(int i, X x) {
        i += n-1;
        dat[i] = x;
        while(i > 0) {
            i = (i-1)/2;
            dat[i] = fx(dat[i*2+1], dat[i*2+2]);
        }
    }
    X output_sub(int a, int b, int k, int l, int r) {
        if(r <= a || b <= l) return ex;
        else if(a <= l && r <= b) return dat[k];
        else {
            X vl = output_sub(a, b, k*2+1, l, (l+r)/2);
            X vr = output_sub(a, b, k*2+2, (l+r)/2, r);
            return fx(vl, vr);
        }
    }
    X output(int a, int b) {
        return output_sub(a, b, 0, 0, n);
    }
};
struct S {
    mat A, B, T;
};
mat matadd(mat A, mat B)
{
    ll N = SZ(A), M = SZ(A[0]);
    REP(i,0,N) {
        REP(j,0,M) A[i][j] += B[i][j], A[i][j] %= MOD;
    }
    return A;
}
mat matmul(mat A, mat B)
{
    ll N = SZ(A), M = SZ(A[0]), K = SZ(B[0]);
    mat C(N, vec(K));
    REP(i,0,N) {
        REP(j,0,K) {
            REP(k,0,M) C[i][j] += A[i][k]*B[k][j], C[i][j] %= MOD;
        }
    }
    return C;
}
int main()
{
    ll N;
    cin >> N;
    mat a(3, vec(1)), b(2, vec(1));
    REP(i,0,3) cin >> a[i][0];
    REP(i,0,2) cin >> b[i][0];
    
    S ex = {mat(3, vec(3)), mat(2, vec(2)), mat(2, vec(3))};
    REP(j,0,3) ex.A[j][j] = 1;
    REP(j,0,2) ex.B[j][j] = 1;
    auto fx = [](S l, S r) -> S {
        S x;
        x.A = matmul(r.A, l.A);
        x.B = matmul(l.B, r.B);
        x.T = matadd(l.T, matmul(l.B, matmul(r.T, l.A)));
        return x;
    };
    SegmentTree<S> st(N+1, fx, ex);
    REP(i,0,N+1) {
        S x = {mat(3, vec(3)), mat(2, vec(2)), mat(2, vec(3))};
        REP(j,0,3) x.A[j][j] = 1;
        REP(j,0,2) x.B[j][j] = 1;
        REP(j,0,2) {
            REP(k,0,3) x.T[j][k] = 6*i+3*j+k;
        }
        st.set(i, x);
    }
    st.build();
    ll Q;
    cin >> Q;
    while(Q--) {
        string t;
        cin >> t;
        if(t == "a") {
            ll i;
            cin >> i;
            S x = st.output(i, i+1);
            REP(j,0,3) {
                REP(k,0,3) cin >> x.A[j][k];
            }
            st.update(i, x);
        }
        else if(t == "b") {
            ll i;
            cin >> i;
            S x = st.output(i, i+1);
            REP(j,0,2) {
                REP(k,0,2) cin >> x.B[j][k];
            }
            st.update(i, x);
        }
        else if(t == "ga") {
            ll i;
            cin >> i;
            S x = st.output(0, i);
            mat ans = matmul(x.A, a);
            cout << ans[0][0] << " " << ans[1][0] << " " << ans[2][0] << endl;
        }
        else if(t == "gb") {
            ll i;
            cin >> i;
            S x = st.output(0, i+1);
            S y = st.output(i+1, N+1);
            mat ans = matmul(y.B, b);
            mat tmp = matmul(y.T, matmul(x.A, a));
            ans = matadd(ans, tmp);
            cout << ans[0][0] << " " << ans[1][0] << endl;
        }
    }
    return 0;
}
/*
*/
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