結果
問題 | No.518 ローマ数字の和 |
ユーザー |
|
提出日時 | 2019-01-26 01:26:07 |
言語 | Nim (2.2.0) |
結果 |
AC
|
実行時間 | 2 ms / 2,000 ms |
コード長 | 1,528 bytes |
コンパイル時間 | 3,909 ms |
コンパイル使用メモリ | 65,552 KB |
実行使用メモリ | 6,944 KB |
最終ジャッジ日時 | 2024-07-01 10:51:54 |
合計ジャッジ時間 | 4,763 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 19 |
ソースコード
import sequtils,strutils,math template times*(n:int,body) = (for _ in 0..<n: body) proc getchar_unlocked():char {. importc:"getchar_unlocked",header: "<stdio.h>" .} proc scan(): int = while true: let k = getchar_unlocked() if k < '0': return result = 10 * result + k.ord - '0'.ord proc parseRoman(S:string): int = return S.replace("CM"," 900 ").replace("CD"," 400 ") .replace("XC"," 90 ").replace("XL"," 40 ") .replace("IX"," 9" ).replace("IV"," 4 ") .replace("M"," 1000 ").replace("D"," 500 ") .replace("C"," 100 ").replace("L"," 50 ") .replace("X"," 10 ").replace("V"," 5 ") .replace("I"," 1 ").replace(" "," ").strip().split().map(parseInt).sum() proc toRoman(n:int):string = if n < 10: if n == 4 : return "IV" if n == 9 : return "IX" if n < 4: return "I".repeat(n) return "V" & "I".repeat(n-5) if n < 100: let left = (n mod 10).toRoman() let my = n div 10 if my == 4 : return "XL" & left if my == 9 : return "XC" & left if my < 4 : return "X".repeat(my) & left return "L" & "X".repeat(my-5) & left if n < 1000: let left = (n mod 100).toRoman() let my = n div 100 if my == 4 : return "CD" & left if my == 9 : return "CM" & left if my < 4 : return "C".repeat(my) & left return "D" & "C".repeat(my-5) & left if n < 4000: let left = (n mod 1000).toRoman() let my = n div 1000 return "M".repeat(my) & left return "ERROR" let n = scan() let r = stdin.readLine.split().map(parseRoman).sum() echo r.toRoman()