結果

問題 No.291 黒い文字列
ユーザー nebukuro09
提出日時 2017-06-10 10:15:25
言語 D
(dmd 2.109.1)
結果
AC  
実行時間 205 ms / 2,000 ms
コード長 2,473 bytes
コンパイル時間 797 ms
コンパイル使用メモリ 114,468 KB
実行使用メモリ 112,068 KB
最終ジャッジ日時 2024-06-12 19:53:15
合計ジャッジ時間 4,679 ms
ジャッジサーバーID
(参考情報)
judge5 / judge2
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 4
other AC * 26
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

import std.stdio, std.array, std.string, std.conv, std.algorithm;
import std.typecons, std.range, std.random, std.math, std.container;
import std.numeric, std.bigint, core.bitop, core.stdc.stdio;
void main() {
auto S = readln.chomp;
auto N = S.length.to!int;
immutable int M = 21;
auto dp = new int[][][][][](N+1, M, M, M, M);
foreach (i; 0..N+1)
foreach (a; 0..M)
foreach (b; 0..M)
foreach (c; 0..M)
fill(dp[i][a][b][c], -1);
dp[0][0][0][0][0] = 0;
foreach (i; 0..N) {
foreach (a; 0..M) {
foreach (b; 0..a+1) {
foreach (c; 0..b+1) {
foreach (d; 0..c+1) {
if (dp[i][a][b][c][d] < 0) continue;
else if (S[i] == 'K' && a < M - 1) dp[i+1][a+1][b][c][d] = max(dp[i][a][b][c][d], dp[i+1][a+1][b][c][d]);
else if (S[i] == 'U' && b < M - 1) dp[i+1][a][b+1][c][d] = max(dp[i][a][b][c][d], dp[i+1][a][b+1][c][d]);
else if (S[i] == 'R' && c < M - 1) dp[i+1][a][b][c+1][d] = max(dp[i][a][b][c][d], dp[i+1][a][b][c+1][d]);
else if (S[i] == 'O' && d < M - 1) dp[i+1][a][b][c][d+1] = max(dp[i][a][b][c][d], dp[i+1][a][b][c][d+1]);
else if (S[i] == 'I' && min(a, b, c, d) > 0) dp[i+1][a-1][b-1][c-1][d-1] = max(dp[i][a][b][c][d] + 1, dp[i+1][a-1][b-1][c-1][d
                            -1]);
else if (S[i] == '?') {
if (a < M - 1) dp[i+1][a+1][b][c][d] = max(dp[i][a][b][c][d], dp[i+1][a+1][b][c][d]);
if (b < M - 1) dp[i+1][a][b+1][c][d] = max(dp[i][a][b][c][d], dp[i+1][a][b+1][c][d]);
if (c < M - 1) dp[i+1][a][b][c+1][d] = max(dp[i][a][b][c][d], dp[i+1][a][b][c+1][d]);
if (d < M - 1) dp[i+1][a][b][c][d+1] = max(dp[i][a][b][c][d], dp[i+1][a][b][c][d+1]);
if (min(a, b, c, d) > 0) dp[i+1][a-1][b-1][c-1][d-1] = max(dp[i][a][b][c][d] + 1, dp[i+1][a-1][b-1][c-1][d-1]);
}
dp[i+1][a][b][c][d] = max(dp[i][a][b][c][d], dp[i+1][a][b][c][d]);
}
}
}
}
}
int ans = 0;
foreach (a; 0..M)
foreach (b; 0..M)
foreach (c; 0..M)
foreach (d; 0..M)
ans = max(ans, dp[N][a][b][c][d]);
ans.writeln;
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0