結果

問題 No.2728 Grid Expansion
ユーザー ___Johniel___Johniel
提出日時 2024-05-04 18:35:24
言語 C++23
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 3,532 bytes
コンパイル時間 2,549 ms
コンパイル使用メモリ 245,660 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-05-04 18:35:28
合計ジャッジ時間 3,478 ms
ジャッジサーバーID
（参考情報） 		judge4 / 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 1 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 1 ms
5,376 KB
testcase_06 AC 1 ms
5,376 KB
testcase_07 AC 1 ms
5,376 KB
testcase_08 AC 2 ms
5,376 KB
testcase_09 AC 1 ms
5,376 KB
testcase_10 AC 1 ms
5,376 KB
testcase_11 AC 1 ms
5,376 KB
testcase_12 AC 1 ms
5,376 KB
testcase_13 AC 2 ms
5,376 KB
testcase_14 AC 1 ms
5,376 KB
testcase_15 AC 2 ms
5,376 KB
testcase_16 AC 1 ms
5,376 KB
testcase_17 AC 2 ms
5,376 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 # 

// github.com/Johniel/contests
// yukicoder/2728/main.cpp

#include <bits/stdc++.h>

#define each(i, c) for (auto& i : c)
#define unless(cond) if (!(cond))
#define makepair(a, b) make_pair(a, b)
// #define endl "\n"

using namespace std;

template<typename P, typename Q> ostream& operator << (ostream& os, pair<P, Q> p);
template<typename P, typename Q> istream& operator >> (istream& is, pair<P, Q>& p);
template<typename P, typename Q, typename R> ostream& operator << (ostream& os, tuple<P, Q, R> t) { os << "(" << get<0>(t) << "," << get<1>(t) << "," << get<2>(t) << ")"; return os; }
template<typename P, typename Q, typename R> istream& operator >> (istream& is, tuple<P, Q, R>& t) { is >> get<0>(t) >> get<1>(t) >> get<2>(t); return is; }
template<typename T> ostream& operator << (ostream& os, vector<T> v) { os << "("; for (auto& i: v) os << i << ","; os << ")"; return os; }
template<typename T> istream& operator >> (istream& is, vector<T>& v) { for (auto& i: v) is >> i; return is; }
template<typename T> ostream& operator << (ostream& os, set<T> s) { os << "set{"; for (auto& i: s) os << i << ","; os << "}"; return os; }
template<typename K, typename V> ostream& operator << (ostream& os, map<K, V> m) { os << "map{"; for (auto& i: m) os << i << ","; os << "}"; return os; }
template<typename E, size_t N> istream& operator >> (istream& is, array<E, N>& a) { for (auto& i: a) is >> i; return is; }
template<typename E, size_t N> ostream& operator << (ostream& os, array<E, N>& a) { os << "[" << N << "]{"; for (auto& i: a) os << i << ","; os << "}"; return os; }
template<typename T> ostream& operator << (ostream& os, stack<T> s) { os << "stack{"; while (s.size()) { os << s.top() << ","; s.pop(); } os << "}"; return os; }
template<typename T> ostream& operator << (ostream& os, queue<T> q) { os << "queue{"; while (q.size()) { os << q.front() << ","; q.pop(); } os << "}"; return os; }
template<typename T> ostream& operator << (ostream& os, deque<T> q) { os << "deque{"; for (int i = 0; i < q.size(); ++i) os << q[i] << ","; os << "}"; return os; }
template<typename T> ostream& operator << (ostream& os, priority_queue<T> q) { os << "heap{"; while (q.size()) { os << q.top() << ","; q.pop(); } os << "}"; return os; }
template<typename P, typename Q> ostream& operator << (ostream& os, pair<P, Q> p) { os << "(" << p.first << "," << p.second << ")"; return os; }
template<typename P, typename Q> istream& operator >> (istream& is, pair<P, Q>& p) { is >> p.first >> p.second; return is; }

template<typename T> inline T setmax(T& a, T b) { return a = std::max(a, b); }
template<typename T> inline T setmin(T& a, T b) { return a = std::min(a, b); }

__attribute__((constructor)) static void _____(void) { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.setf(ios_base::fixed); cout.precision(15); return ; }

using lli = long long int;
using ull = unsigned long long;
using str = string;
template<typename T> using vec = vector<T>;

// constexpr lli mod = 1e9 + 7;
constexpr lli mod = 998244353;

int main(int argc, char *argv[])
{
  int n, k;
  while (cin >> n >> k) {
    char g[n*k][n*k];
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        char c;
        cin >> c;
        for (int a = 0; a < k; ++a) {
          for (int b = 0; b < k; ++b) {
            g[i * k + a][j * k + b] = c;
          }
        }
      }
    }
    for (int i = 0; i < n * k; ++i) {
      for (int j = 0; j < n * k; ++j) {
        cout << g[i][j];
      }
      cout << endl;
    }
  }
  return 0;
}
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