結果

問題 No.493 とても長い数列と文字列(Long Long Sequence and a String）
ユーザー Pachicobue
提出日時 2017-08-13 20:03:06
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,686 bytes
コンパイル時間 1,565 ms
コンパイル使用メモリ 175,808 KB
実行使用メモリ 6,824 KB
最終ジャッジ日時 2024-10-13 09:14:24
合計ジャッジ時間 4,367 ms
ジャッジサーバーID
（参考情報） 		judge4 / 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 * 4
other AC * 55 WA * 60
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

// {{{ Templates
#include <bits/stdc++.h>
#define show(x) cerr << #x << " = " << x << endl
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{
    os << "sz:" << v.size() << "\n[";
    for (const auto& p : v) {
        os << p << ",";
    }
    os << "]\n";
    return os;
}
template <typename S, typename T>
ostream& operator<<(ostream& os, const pair<S, T>& p)
{
    os << "(" << p.first << "," << p.second
       << ")";
    return os;
}
constexpr ll MOD = (ll)1e9 + 7LL;
template <typename T>
constexpr T INF = numeric_limits<T>::max() / 100;
// }}}
using pll = pair<ll, ll>;
inline ll convert(const char c)
{
    return c == '0' ? 10 : c - '0';
}
inline pll calcst(const string& s, const ll l, const ll r)
{
    assert(0 <= l);
    assert(l < r);
    assert(r <= s.size());
    ll sum = 0;
    ll prod = 1;
    for (ll i = l; i < r; i++) {
        const char c = s[i];
        sum = (sum + convert(c)) % MOD;
        prod = (prod * convert(c)) % MOD;
    }
    return make_pair(sum, prod);
}
pll rec(const ll n, const ll l, const ll r, const vector<ll>& length, const vector<pll>& calced)  // f(n)の[L,R)の(和,積)
{
    assert(l >= 0);
    assert(l < r);
    if (n == 0) {
        return make_pair(0, 1);
    }
    if (l == 0 and r == length[n]) {
        return calced[n];
    } else {
        const string intst = to_string(n * n);
        const ll interval = intst.size();
        const ll half = (length[n] - interval) / 2;
        pll ret = make_pair(0, 1);
        if (l < half) {
            const pll p = rec(n - 1, l, min(r, half), length, calced);
            ret.first = (ret.first + p.first) % MOD;
            ret.second = (ret.second * p.second) % MOD;
        }
        if (r > half + interval) {
            const pll p = rec(n - 1, max(l - half - interval, 0LL), r - half - interval, length, calced);
            ret.first = (ret.first + p.first) % MOD;
            ret.second = (ret.second * p.second) % MOD;
        }
        if (r > half and l < half + interval) {
            const pll p = calcst(intst, max(l - half, 0LL), min(interval, r - half));
            ret.first = (ret.first + p.first) % MOD;
            ret.second = (ret.second * p.second) % MOD;
        }
        return ret;
    }
}
inline pll cal(const string& s)
{
    ll sum = 0;
    ll prod = 1;
    for (const char c : s) {
        sum += convert(c);
        sum = sum % MOD;
        prod *= convert(c);
        prod = prod % MOD;
    }
    return make_pair(sum, prod);
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll k, l, r;
    cin >> k >> l >> r;
    l--, r--;
    constexpr ll STOP = (ll)1e18;
    vector<ll> length(1, 0);
    ll len = 0;
    for (ll i = 1;; i++) {
        len = len * 2 + (ll)log10(i * i) + 1;
        length.push_back(len);
        if (len > STOP) {
            break;
        }
    }
    const int size = length.size();
    if (k > size) {
        k = size;
    }
    vector<pll> calced(size, make_pair(0, 1));
    for (int i = 1; i < size; i++) {
        const string s = to_string(i * i);
        const pll p = cal(s);
        calced[i].first = calced[i - 1].first * 2 + p.first;
        calced[i].second = ((calced[i - 1].second * calced[i - 1].second) % MOD) * p.second;
        calced[i].first = calced[i].first % MOD;
        calced[i].second = calced[i].second % MOD;
    }
    // show(calced);
    if (length[k] <= r) {
        cout << -1 << endl;
        return 0;
    }
    const pll ans = rec(k, l, r + 1, length, calced);
    cout << ans.first << " " << ans.second << endl;
    return 0;
}
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