C++の練習を兼ねて, AtCoder Grand Contest 041 の 問題C (Domino Quality) を解いてみた.
■感想.
1. 問題Cは, 解答方針が見えなかったので, 解説を参考に実装して, ようやく, AC版に到達できた.
2. ドミノ牌の敷き詰め方に, 不思議な性質があることを知って, 個人的には, 非常に面白い問題に感じた.
3. 時間を見つけて, 引き続き, 過去問を振り返っていきたいと思う.
本家のサイト AtCoder Grand Contest 041 解説 を ご覧下さい.
■C++版プログラム(問題C/AC版).
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// 解き直し. // https://img.atcoder.jp/agc041/editorial.pdf // C++(GCC 9.2.1) #include <bits/stdc++.h> using namespace std; #define repex(i, a, b, c) for(int i = a; i < b; i += c) #define repx(i, a, b) repex(i, a, b, 1) #define rep(i, n) repx(i, 0, n) #define repr(i, a, b) for(int i = a; i >= b; i--) char ans[1010][1010]; char q31[3][3] = { {'a', 'a', '.'}, {'.', '.', 'a'}, {'.', '.', 'a'} }; char q43[4][4] = { {'a', 'a', 'b', 'c'}, {'d', 'd', 'b', 'c'}, {'b', 'c', 'a', 'a'}, {'b', 'c', 'd', 'd'} }; char q53[5][5] = { {'a', 'a', 'b', 'b', 'a'}, {'b', 'c', 'c', '.', 'a'}, {'b', '.', '.', 'c', 'b'}, {'a', '.', '.', 'c', 'b'}, {'a', 'b', 'b', 'a', 'a'} }; char q63[6][6] = { {'a', 'a', 'b', 'c', '.', '.'}, {'d', 'd', 'b', 'c', '.', '.'}, {'.', '.', 'a', 'a', 'b', 'c'}, {'.', '.', 'd', 'd', 'b', 'c'}, {'b', 'c', '.', '.', 'a', 'a'}, {'b', 'c', '.', '.', 'd', 'd'} }; char q73[7][7] = { {'a', 'a', 'b', 'b', 'c', 'c', '.'}, {'d', 'd', '.', 'd', 'd', '.', 'a'}, {'.', '.', 'd', '.', '.', 'd', 'a'}, {'.', '.', 'd', '.', '.', 'd', 'b'}, {'d', 'd', '.', 'd', 'd', '.', 'b'}, {'.', '.', 'd', '.', '.', 'd', 'c'}, {'.', '.', 'd', '.', '.', 'd', 'c'} }; int main(){ // 1. 入力情報. int N; scanf("%d", &N); // 2. N = 2. if(N == 2){ puts("-1"); return 0; } // 3. N = 3. if(N == 3){ rep(i, 3){ rep(j, 3) printf("%c", q31[i][j]); puts(""); } return 0; } // 4. N >= 4. // 4-1. x, y を 算出. int x = N / 4, y = N - x * 4, s, e; if(y < 4) x--, y += 4; rep(i, N) rep(j, N) ans[i][j] = '.'; // 4-2. 4 * 4 の マス(x個). rep(k, x){ // 左上が, k番目の場合, (4 * k, 4 * k) ~ (4 * k + 3, 4 * k + 3) // の 範囲 を, q43 で 埋めるようにする. s = 4 * k; e = 4 * (k + 1); repx(i, s, e){ int di = i - s; repx(j, s, e){ int dj = j - s; ans[i][j] = q43[di][dj]; } } } // 4-3. y * y の マス(1個). s = 4 * x; e = N; repx(i, s, e){ int di = i - s; repx(j, s, e){ int dj = j - s; if(y == 4) ans[i][j] = q43[di][dj]; if(y == 5) ans[i][j] = q53[di][dj]; if(y == 6) ans[i][j] = q63[di][dj]; if(y == 7) ans[i][j] = q73[di][dj]; } } // 5. 出力. rep(i, N){ rep(j, N) printf("%c", ans[i][j]); puts(""); } return 0; } |
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[入力例] 6 [出力例] aabb.. b..zz. ba.... .a..aa ..a..b ..a..b ※AtCoderテストケースより ※但し, 上記のプログラムでは, 以下の内容が出力される. aabc.. ddbc.. ..aabc ..ddbc bc..aa bc..dd [入力例] 2 [出力例] -1 ※AtCoderテストケースより [入力例] 3 [出力例] aa. ..a ..a [入力例] 7 [出力例] aabbcc. dd.dd.a ..d..da ..d..db dd.dd.b ..d..dc ..d..dc [入力例] 8 [出力例] aabc.... ddbc.... bcaa.... bcdd.... ....aabc ....ddbc ....bcaa ....bcdd [入力例] 9 [出力例] aabc..... ddbc..... bcaa..... bcdd..... ....aabba ....bcc.a ....b..cb ....a..cb ....abbaa [入力例] 10 [出力例] aabc...... ddbc...... bcaa...... bcdd...... ....aabc.. ....ddbc.. ......aabc ......ddbc ....bc..aa ....bc..dd [入力例] 11 [出力例] aabc....... ddbc....... bcaa....... bcdd....... ....aabbcc. ....dd.dd.a ......d..da ......d..db ....dd.dd.b ......d..dc ......d..dc [入力例] 33 [出力例] aabc............................. ddbc............................. bcaa............................. bcdd............................. ....aabc......................... ....ddbc......................... ....bcaa......................... ....bcdd......................... ........aabc..................... ........ddbc..................... ........bcaa..................... ........bcdd..................... ............aabc................. ............ddbc................. ............bcaa................. ............bcdd................. ................aabc............. ................ddbc............. ................bcaa............. ................bcdd............. ....................aabc......... ....................ddbc......... ....................bcaa......... ....................bcdd......... ........................aabc..... ........................ddbc..... ........................bcaa..... ........................bcdd..... ............................aabba ............................bcc.a ............................b..cb ............................a..cb ............................abbaa [入力例] 55 [出力例] aabc................................................... ddbc................................................... bcaa................................................... bcdd................................................... ....aabc............................................... ....ddbc............................................... ....bcaa............................................... ....bcdd............................................... ........aabc........................................... ........ddbc........................................... ........bcaa........................................... ........bcdd........................................... ............aabc....................................... ............ddbc....................................... ............bcaa....................................... ............bcdd....................................... ................aabc................................... ................ddbc................................... ................bcaa................................... ................bcdd................................... ....................aabc............................... ....................ddbc............................... ....................bcaa............................... ....................bcdd............................... ........................aabc........................... ........................ddbc........................... ........................bcaa........................... ........................bcdd........................... ............................aabc....................... ............................ddbc....................... ............................bcaa....................... ............................bcdd....................... ................................aabc................... ................................ddbc................... ................................bcaa................... ................................bcdd................... ....................................aabc............... ....................................ddbc............... ....................................bcaa............... ....................................bcdd............... ........................................aabc........... ........................................ddbc........... ........................................bcaa........... ........................................bcdd........... ............................................aabc....... ............................................ddbc....... ............................................bcaa....... ............................................bcdd....... ................................................aabbcc. ................................................dd.dd.a ..................................................d..da ..................................................d..db ................................................dd.dd.b ..................................................d..dc ..................................................d..dc |
■参照サイト
AtCoder Grand Contest 041