結果
問題 | No.488 四角関係 |
ユーザー |
👑 |
提出日時 | 2019-04-08 11:28:07 |
言語 | Lua (LuaJit 2.1.1734355927) |
結果 |
AC
|
実行時間 | 167 ms / 5,000 ms |
コード長 | 874 bytes |
コンパイル時間 | 89 ms |
コンパイル使用メモリ | 5,376 KB |
実行使用メモリ | 5,376 KB |
最終ジャッジ日時 | 2024-07-01 22:50:23 |
合計ジャッジ時間 | 1,409 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 22 |
ソースコード
tall = {}function check(i1, i2, i3, i4)if(tall[i1][i2] and tall[i1][i3] and tall[i2][i4] and tall[i3][i4] and (tall[i1][i4] == nil and tall[i2][i3] == nil)) thenreturn trueelseif(tall[i1][i3] and tall[i1][i4] and tall[i2][i3] and tall[i2][i4] and (tall[i1][i2] == nil and tall[i3][i4] == nil)) thenreturn trueelseif(tall[i1][i2] and tall[i1][i4] and tall[i2][i3] and tall[i3][i4] and (tall[i1][i3] == nil and tall[i2][i4] == nil)) thenreturn trueendreturn falseendn, m = io.read("*n", "*n")for i = 1, n dotall[i] = {}endfor i = 1, m doa, b = io.read("*n", "*n")a, b = a + 1, b + 1tall[a][b] = trueendcnt = 0for i_a = 1, n - 3 dofor i_b = i_a + 1, n - 2 dofor i_c = i_b + 1, n - 1 dofor i_d = i_c + 1, n doif(check(i_a, i_b, i_c, i_d)) then cnt = cnt + 1 endendendendendprint(cnt)