結果

問題 No.1172 Add Recursive Sequence
ユーザー tokusakuraitokusakurai
提出日時 2020-08-14 23:08:06
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 454 ms / 4,000 ms
コード長 4,137 bytes
コンパイル時間 2,658 ms
コンパイル使用メモリ 204,444 KB
実行使用メモリ 6,740 KB
最終ジャッジ日時 2024-04-18 23:01:16
合計ジャッジ時間 4,774 ms
ジャッジサーバーID
(参考情報)
judge2 / judge5
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
5,248 KB
testcase_01 AC 2 ms
5,376 KB
testcase_02 AC 2 ms
5,376 KB
testcase_03 AC 2 ms
5,376 KB
testcase_04 AC 2 ms
5,376 KB
testcase_05 AC 2 ms
5,376 KB
testcase_06 AC 5 ms
5,376 KB
testcase_07 AC 5 ms
5,376 KB
testcase_08 AC 5 ms
5,376 KB
testcase_09 AC 5 ms
5,376 KB
testcase_10 AC 34 ms
5,376 KB
testcase_11 AC 36 ms
5,376 KB
testcase_12 AC 37 ms
5,376 KB
testcase_13 AC 36 ms
5,376 KB
testcase_14 AC 223 ms
6,612 KB
testcase_15 AC 218 ms
6,740 KB
testcase_16 AC 454 ms
6,620 KB
testcase_17 AC 446 ms
6,492 KB
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 #

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, x, n) for(int i = x; i <= n; i++)
#define rep3(i, x, n) for(int i = x; i >= n; i--)
#define elif else if
#define sp(x) fixed << setprecision(x)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const ll MOD = 1e9+7;
//const ll MOD = 998244353;
const int inf = (1<<30)-1;
const ll INF = (1LL<<60)-1;
const ld EPS = 1e-10;
template<typename T> bool chmax(T &x, const T &y) {return (x < y)? (x = y, true) : false;};
template<typename T> bool chmin(T &x, const T &y) {return (x > y)? (x = y, true) : false;};

template<ll mod>
struct Mod_Int{
    ll x;
    Mod_Int() {}
    Mod_Int(ll y) : x (y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

    Mod_Int &operator += (const Mod_Int &p){
        x = (x + p.x) % mod;
        return *this;
    }

    Mod_Int &operator -= (const Mod_Int &p){
        x = (x + mod - p.x) % mod;
        return *this;
    }

    Mod_Int &operator *= (const Mod_Int &p){
        x = (x * p.x) % mod;
        return *this;
    }

    Mod_Int &operator /= (const Mod_Int &p){
        *this *= p.inverse();
        return *this;
    }

    Mod_Int &operator ++ () {return *this += Mod_Int(1);}

    Mod_Int operator ++ (int){
        Mod_Int tmp = *this;
        ++*this;
        return tmp;
    }

    Mod_Int &operator -- () {return *this -= Mod_Int(1);}

    Mod_Int operator -- (int){
        Mod_Int tmp = *this;
        --*this;
        return tmp;
    }

    Mod_Int operator - () const {return Mod_Int(-x);}

    Mod_Int operator + (const Mod_Int &p) const {return Mod_Int(*this) += p;}

    Mod_Int operator - (const Mod_Int &p) const {return Mod_Int(*this) -= p;}

    Mod_Int operator * (const Mod_Int &p) const {return Mod_Int(*this) *= p;}

    Mod_Int operator / (const Mod_Int &p) const {return Mod_Int(*this) /= p;}

    bool operator == (const Mod_Int &p) const {return x == p.x;}

    bool operator != (const Mod_Int &p) const {return x != p.x;}

    Mod_Int pow(ll n) const{
        Mod_Int now = *this, ret = 1;
        while(n > 0){
            if(n & 1) ret *= now;
            now *= now, n >>= 1;
        }
        return ret;
    }

    Mod_Int inverse() const{
        return pow(mod-2);
    }

    friend ostream &operator << (ostream &os, const Mod_Int &p){
        return os << p.x;
    }

    friend istream &operator >> (istream &is, Mod_Int &p){
        ll a;
        is >> a;
        p = Mod_Int<mod>(a);
        return is;
    }
};

using mint = Mod_Int<MOD>;
const int MAX_N = 1e6;
mint fac[MAX_N+1], ifac[MAX_N+1];

void init(){
    fac[0] = 1;
    rep2(i, 1, MAX_N){
        fac[i] = fac[i-1]*i;
    }
    ifac[MAX_N] = fac[MAX_N].inverse();
    rep3(i, MAX_N, 1){
        ifac[i-1] = ifac[i]*i;
    }
}

mint comb(int n, int k){
    return fac[n]*ifac[n-k]*ifac[k];
}

mint perm(int n, int k){
    return fac[n]*ifac[n-k];
}

int main(){
    int K, N, M;
    cin >> K >> N >> M;
    mint a[K], c[K];
    rep(i, K) cin >> a[i];
    rep(i, K) cin >> c[i];
    mint b[N];
    rep(i, N){
        if(i < K) b[i] = a[i];
        else{
            b[i] = 0;
            rep(j, K) b[i] += c[j]*b[i-1-j];
        }
    }
    vector<pii> p;
    rep(i, M){
        int l, r; cin >> l >> r;
        p.eb(l, 0), p.eb(r, r-l);
    }
    sort(all(p));
    int pos = K;
    mint ans[N];
    fill(ans, ans+N, 0);
    for(auto &e: p){
        int nxt = e.first+K;
        rep2(i, pos, min(N, nxt)-1){
            rep(j, K) ans[i] += c[j]*ans[i-1-j];
        }
        pos = nxt;
        if(e.second == 0){
            rep2(i, e.first, min(e.first+K, N)-1){
                ans[i] += a[i-e.first];
            }
        }
        else{
            rep2(i, e.first, min(e.first+K, N)-1){
                ans[i] -= b[e.second+i-e.first];
            }
        }
    }
    rep2(i, pos, N-1){
        rep(j, K) ans[i] += c[j]*ans[i-1-j];
    }
    rep(i, N) cout << ans[i] << endl;
}
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