結果

問題 No.1515 Making Many Multiples
ユーザー gazellegazelle
提出日時 2021-05-21 22:05:58
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
TLE  
実行時間 -
コード長 6,501 bytes
コンパイル時間 2,973 ms
コンパイル使用メモリ 220,324 KB
実行使用メモリ 43,392 KB
最終ジャッジ日時 2024-10-10 08:49:26
合計ジャッジ時間 41,840 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge4
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 4 ms
11,136 KB
testcase_01 AC 1,837 ms
42,752 KB
testcase_02 TLE -
testcase_03 TLE -
testcase_04 TLE -
testcase_05 TLE -
testcase_06 AC 1,893 ms
43,264 KB
testcase_07 AC 1,339 ms
43,264 KB
testcase_08 TLE -
testcase_09 TLE -
testcase_10 TLE -
testcase_11 TLE -
testcase_12 TLE -
testcase_13 WA -
testcase_14 TLE -
testcase_15 TLE -
testcase_16 TLE -
testcase_17 RE -
testcase_18 RE -
testcase_19 RE -
testcase_20 RE -
testcase_21 WA -
testcase_22 WA -
testcase_23 WA -
testcase_24 AC 5 ms
11,136 KB
testcase_25 AC 4 ms
10,944 KB
testcase_26 AC 5 ms
11,104 KB
testcase_27 AC 1,600 ms
42,552 KB
testcase_28 RE -
testcase_29 AC 40 ms
12,544 KB
testcase_30 AC 823 ms
36,668 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>
#define FOR(i, n, m) for(ll i = (n); i < (ll)(m); i++)
#define REP(i, n) FOR(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define pb push_back
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<ll, ll>;
constexpr ll inf = 1000000000;
constexpr ll mod = 998244353;
constexpr long double eps = 1e-6;
 
template<typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> p) {
    os << to_string(p.first) << " " << to_string(p.second);
    return os;
}
template<typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
    REP(i, v.size()) {
        if(i) os << " ";
        os << v[i];
    }
    return os;
}
 
struct modint {
    ll n;
public:
    modint(const ll n = 0) : n((n % mod + mod) % mod) {}
    static modint pow(modint a, int m) {
        modint r = 1;
        while(m > 0) {
            if(m & 1) { r *= a; }
            a = (a * a); m /= 2;
        }
        return r;
    }
    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; }
    modint operator~() const { return (this -> pow(n, mod - 2)); } // inverse
    friend bool operator==(const modint& lhs, const modint& rhs) {
        return lhs.n == rhs.n;
    }
    friend bool operator<(const modint& lhs, const modint& rhs) {
        return lhs.n < rhs.n;
    }
    friend bool operator>(const modint& lhs, const modint& rhs) {
        return lhs.n > rhs.n;
    }
    friend modint &operator+=(modint& lhs, const modint& rhs) {
        lhs.n += rhs.n;
        if (lhs.n >= mod) lhs.n -= mod;
        return lhs;
    }
    friend modint &operator-=(modint& lhs, const modint& rhs) {
        lhs.n -= rhs.n;
        if (lhs.n < 0) lhs.n += mod;
        return lhs;
    }
    friend modint &operator*=(modint& lhs, const modint& rhs) {
        lhs.n = (lhs.n * rhs.n) % mod;
        return lhs;
    }
    friend modint &operator/=(modint& lhs, const modint& rhs) {
        lhs.n = (lhs.n * (~rhs).n) % mod;
        return lhs;
    }
    friend modint operator+(const modint& lhs, const modint& rhs) {
        return modint(lhs.n + rhs.n);
    }
    friend modint operator-(const modint& lhs, const modint& rhs) {
        return modint(lhs.n - rhs.n);
    }
    friend modint operator*(const modint& lhs, const modint& rhs) {
        return modint(lhs.n * rhs.n);
    }
    friend modint operator/(const modint& lhs, const modint& rhs) {
        return modint(lhs.n * (~rhs).n);
    }
};
istream& operator>>(istream& is, modint m) { is >> m.n; return is; }
ostream& operator<<(ostream& os, modint m) { os << m.n; return os; }
 
#define MAX_N 1010101
long long extgcd(long long a, long long b, long long& x, long long& y) {
    long long d = a;
    if (b != 0) {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    } else {
        x = 1; y = 0;
    }
    return d;
}
long long mod_inverse(long long a, long long m) {
    long long x, y;
    if(extgcd(a, m, x, y) == 1) return (m + x % m) % m;
    else return -1;
}
vector<long long> fact(MAX_N+1, inf);
long long mod_fact(long long n, long long& e) {
    if(fact[0] == inf) {
        fact[0]=1;
        if(MAX_N != 0) fact[1]=1;
        for(ll i = 2; i <= MAX_N; ++i) {
            fact[i] = (fact[i-1] * i) % mod;
        }
    }
    e = 0;
    if(n == 0) return 1;
    long long res = mod_fact(n / mod, e);
    e += n / mod;
    if((n / mod) % 2 != 0) return (res * (mod - fact[n % mod])) % mod;
    return (res * fact[n % mod]) % mod;
}
// return nCk
long long mod_comb(long long n, long long k) {
    if(n < 0 || k < 0 || n < k) return 0;
    long long e1, e2, e3;
    long long a1 = mod_fact(n, e1), a2 = mod_fact(k, e2), a3 = mod_fact(n - k, e3);
    if(e1 > e2 + e3) return 0;
    return (a1 * mod_inverse((a2 * a3) % mod, mod)) % mod;
}
 
using mi = modint;
mi mod_pow(mi a, ll n) {
    mi ret = 1;
    mi tmp = a;
    while(n > 0) {
        if(n % 2) ret *= tmp;
        tmp = tmp * tmp;
        n /= 2;
    }
    return ret;
}

ll mod_pow(ll a, ll n, ll m) {
    ll ret = 1;
    ll tmp = a;
    while(n > 0) {
        if(n % 2) ret *= tmp;
        ret %= m;
        tmp = tmp * tmp;
        tmp %= m;
        n /= 2;
    }
    return ret % m;
}
 
ll gcd(ll a, ll b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

template <typename T> struct segment_tree {
private:
	int n;
	const function<T(T, T)> op;
	const T ie;
	vector<T> seq;

public:
	segment_tree(int _n, function<T(T, T)> op, const T ie) : op(op), ie(ie) {
		n = 1;
		while(n < _n) n *= 2;
		seq.resize(2 * n - 1);
		for(int i = 0; i < 2 * n - 1; i++) seq[i] = ie;
	}

	void update(int k, const T e) {
		k += n - 1;
		seq[k] = e;
		while(k > 0) {
			k = (k - 1) / 2;
			seq[k] = op(seq[k * 2 + 1], seq[k * 2 + 2]);
		}
	}

	T get(int k) {
		k += n - 1;
		return seq[k];
	}

	T query(int a, int b, int k = 0, int l = 0, int r = -1) {
		if(r == -1) r = n;
		if(r <= a || b <= l) return ie;
		if(a <= l && r <= b) return seq[k];
		T vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
		T vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
		return op(vl, vr);
	}
};
// segment_tree<ll> rsq(n, [] (ll a, ll b) { return a + b; }, 0)
// segment_tree<ll> rMq(n, [] (ll a, ll b) { return max(a, b); }, -1e18)
// segment_tree<ll> rmq(n, [] (ll a, ll b) { return min(a, b); }, 1e18)

int a[2002];
int n, k, x, y;

int solve() {
    vector<segment_tree<int>> segs(n, segment_tree<int>(n, [] (int a, int b) { return max(a, b); }, -1e9));

    segs[1].update((x + y) % k, 0);

    for(int i = 2; i < n; ++i) {
        int rev = (k - a[i]) % k;
        for(int j = i - 1; j >= 1; --j) {
            if(segs[j].query(rev, rev + 1) != -1e9) {
                segs[j].update(rev, segs[j].query(rev, rev + 1) + 1);
            }
        }
        for(int j = i - 1; j >= 1; --j) {
            int sum = (a[i] + a[j]) % k;
            int crt = segs[i].query(sum, sum + 1);
            int next = segs[j].query(0, k);
            segs[i].update(sum, max(crt, next));
        }
    }

    int ans = segs[n - 1].query(0, k);
    return ans;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> n >> k >> x >> y;

    a[0] = x;
    a[1] = y;
    REP(i, n) cin >> a[2 + i];
    REP(i, n + 2) a[i] %= k;

    n += 2;

    int ans = solve();
    swap(a[0], a[1]);
    ans = max(ans, solve());

    cout << ans << endl;

    return 0;
}
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