結果

問題 No.1097 Remainder Operation
ユーザー hitonanodehitonanode
提出日時 2020-06-26 21:32:36
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 143 ms / 2,000 ms
コード長 5,529 bytes
コンパイル時間 2,270 ms
コンパイル使用メモリ 209,996 KB
実行使用メモリ 51,500 KB
最終ジャッジ日時 2024-07-04 19:48:54
合計ジャッジ時間 5,378 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
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 2 ms
5,376 KB
testcase_07 AC 11 ms
7,936 KB
testcase_08 AC 11 ms
7,936 KB
testcase_09 AC 10 ms
8,064 KB
testcase_10 AC 10 ms
8,064 KB
testcase_11 AC 11 ms
7,936 KB
testcase_12 AC 118 ms
51,368 KB
testcase_13 AC 135 ms
51,376 KB
testcase_14 AC 139 ms
51,496 KB
testcase_15 AC 143 ms
51,372 KB
testcase_16 AC 125 ms
51,352 KB
testcase_17 AC 85 ms
51,372 KB
testcase_18 AC 74 ms
51,376 KB
testcase_19 AC 97 ms
51,500 KB
testcase_20 AC 100 ms
51,500 KB
testcase_21 AC 139 ms
51,372 KB
testcase_22 AC 140 ms
51,368 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;
using lint = long long;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template <typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template <typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template <typename V, typename T> void ndfill(V &x, const T &val) { x = val; }
template <typename V, typename T> void ndfill(vector<V> &vec, const T &val) { for (auto &v : vec) ndfill(v, val); }
template <typename T> bool chmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template <typename T> bool chmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template <typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template <typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
template <typename T> vector<T> srtunq(vector<T> vec) { sort(vec.begin(), vec.end()), vec.erase(unique(vec.begin(), vec.end()), vec.end()); return vec; }
template <typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto &v : vec) is >> v; return is; }
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template <typename T> ostream &operator<<(ostream &os, const deque<T> &vec) { os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template <typename T> ostream &operator<<(ostream &os, const set<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template <typename T> ostream &operator<<(ostream &os, const multiset<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa) { os << "(" << pa.first << "," << pa.second << ")"; return os; }
template <typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp) { os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template <typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp) { os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;

// Binary lifting / `Doubling`
// Complexity: O(NlogN) precalculation / O(logN) per query
// <https://atcoder.jp/contests/arc060/submissions/7039451>
struct BinaryLifting
{
    int N, INVALID, lgD;
    std::vector<std::vector<int>> mat;
    BinaryLifting() : N(0), lgD(0) {}
    BinaryLifting(const std::vector<int> &vec_nxt, int INVALID = -1, int lgd = 0) : N(vec_nxt.size()), INVALID(INVALID), lgD(lgd)
    {
        while ((1LL << lgD) < N) lgD++;
        mat.assign(lgD, std::vector<int>(N, INVALID));
        mat[0] = vec_nxt;
        for (int i = 0; i < N; i++) if (mat[0][i] < 0 or mat[0][i] >= N) mat[0][i] = INVALID;
        for (int d = 0; d < lgD - 1; d++) {
            for (int i = 0; i < N; i++) if (mat[d][i] != INVALID) mat[d + 1][i] = mat[d][mat[d][i]];
        }
    }
    int kth_next(int now, long long k)
    {
        if (k >= (1LL << lgD)) exit(8);
        for (int d = 0; k and now != INVALID; d++, k >>= 1) if (k & 1) now = mat[d][now];
        return now;
    }

    // Distance from l to [r, \infty)
    // Requirement: mat[0][i] > i for all i (monotone increasing)
    int distance(int l, int r)
    {
        if (l >= r) return 0;
        int ret = 0;
        for (int d = lgD - 1; d >= 0; d--) {
            if (mat[d][l] < r and mat[d][l] != INVALID) ret += 1 << d, l = mat[d][l];
        }
        if (mat[0][l] == INVALID or mat[0][l] >= r) return ret + 1;
        else return -1; // Unable to reach
    }
};
int main()
{
    int N;
    cin >> N;
    vector<int> A(N);
    cin >> A;

    int D = 40;
    vector<vector<int>> nxt(D, vector<int>(N));
    vector<vector<lint>> d(D, vector<lint>(N));
    REP(i, nxt[0].size())
    {
        nxt[0][i] = (i + A[i % N]) % N;
        d[0][i] = A[i % N];
    }
    FOR(j, 1, D)
    {
        REP(i, N) nxt[j][i] = nxt[j - 1][nxt[j - 1][i]];
        REP(i, N) d[j][i] = d[j - 1][i] + d[j - 1][nxt[j - 1][i]];
    }
    int Q;
    cin >> Q;
    while (Q--)
    {
        lint k;
        cin >> k;
        int now = 0;
        lint ret = 0;
        REP(j, D)
        {
            if ((k >> j) & 1)
            {
                ret += d[j][now];
                now = nxt[j][now];
            }
        }
        cout << ret << '\n';
    }
}
0