結果

問題 No.991 N×Mマス計算（構築）
コンテスト
ユーザー fine
提出日時 2020-02-15 01:16:59
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 19 ms / 2,000 ms
コード長 2,485 bytes
コンパイル時間 1,671 ms
コンパイル使用メモリ 170,620 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-10-06 13:35:36
合計ジャッジ時間 4,568 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
（要ログイン）
ファイルパターン 結果
sample AC * 2
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 # 

#include <bits/stdc++.h>

using namespace std;

using ll = long long;

ll calc(ll x) {
    ll ok = 1, ng = x + 1;
    while (ng - ok > 1) {
        ll mid = (ok + ng) / 2;
        (mid * mid <= x ? ok : ng) = mid;
    }
    return ok;
}

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

    ll n = 100000, m = 100000;
    ll K = 1000000000;
    char op = '*';

    ll x;
    cin >> x;

    if (x == 0) {
        cout << "3 2 129\n";
        cout << "+ 12 34\n";
        cout << 10 << endl;
        cout << 20 << endl;
        cout << 30 << endl;
        return 0;
    }

    ll hoge = calc(x);
    ll rest = x - hoge * hoge;
    ll a1 = hoge, a2 = 0, b1 = 0, b2 = hoge;
    if (rest > 0) {
        ll diff = 1e15;
        for (ll i = 1; i * i <= rest && i <= hoge; ++i) {
            if (rest % i != 0) continue;
            ll j = rest / i;

            if (abs(i - j) < diff) {
                a1 = i;
                a2 = hoge - i;
                b1 = j;
                diff = abs(i - j);
            }
        }
    }

    vector<ll> a(n), b(m);
    ll asum = 0, bsum = 0;
    for (ll i = 0; i < a1; ++i) {
        a[i] = 50000;
        (asum += a[i]) %= K;
    }
    for (ll i = 0; i < a2; ++i) {
        a[i + a1] = 25000;
        (asum += a[i + a1]) %= K;
    }
    for (ll i = 0; i < b1; ++i) {
        b[i] = 20000;
        (bsum += b[i]) %= K;
    }
    for (ll i = 0; i < b2; ++i) {
        b[i + b1] = 40000;
        (bsum += b[i + b1]) %= K;
    }
    ll rb = (K + 1 - bsum) % K;
    ll ra = (K + x - asum) % K;

    ll i = a1 + a2;
    ll numa = n - i;
    if (ra < numa) ra += K;
    ll vala = ra / numa;
    for (ll ii = 0; ii < numa; ++ii) {
        a[i + ii] = vala;
        if (ii < ra % numa) a[i + ii]++;
        (asum += a[i + ii]) %= K;
    }

    ll j = b1 + b2;
    ll numb = m - j;
    if (rb < numb) rb += K;
    ll valb = rb / numb;
    for (ll ii = 0; ii < numb; ++ii) {
        b[j + ii] = valb;
        if (ii < rb % numb) b[j + ii]++;
        (bsum += b[j + ii]) %= K;
    }
    assert(asum * bsum % K == x);
    cerr << vala << " " << valb << endl;

    cout << n << " " << m << " " << K << "\n";
    cout << op;
    for (ll i = 0; i < m; ++i) {
        cout << " " << b[i]; 
    }
    cout << "\n";
    for (int i = 0; i < n; ++i) {
        cout << a[i] << "\n";
    }
    return 0;
}

// (sb * n + sa * m) % K == X
// K = 10^9
// sb = pb + rb, sa = pa + ra
// (rb * n + ra * m) % K == (X - pb * n - pa * m) % K 

// sa * sb == X
0