[BOJ] 3101번 토끼의 이동

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코드

• C++ 17

#include <cstdio>   using namespace std;   long long Sigma(int N) {     return 1LL * N * (N + 1LL) / 2LL; }   int main() {     int N, K, R = 0, C = 0;     scanf("%d %d\n", &N, &K);     long long result = 1;     for (int i = 0; i < K; i++)     {         switch (getchar())         {         case 'U':             --R;             break;         case 'D':             ++R;             break;         case 'L':             --C;             break;         default:             ++C;         }                  // 좌표를 이용해서 값을 얻어낼 수 있다.         long long first, dist;         if (R + C < N)         {             // 각 줄은 1의 등차수열이다.             // 첫번째 값을 얻어내고             first = 1LL + Sigma(R + C);             // 첫번쨰 값의 좌표로부터 얼마나 떨어져 있는지를 계산한다.             dist = ((R + C) % 2 ? R : C);         }          // 여기를 계산할 때 주의해야 한다.         else         {             first = 1LL + Sigma(N) + Sigma(N - 1) - Sigma(2 * N - (R + C + 1));             dist = ((R + C) % 2 ? R : C) - ((long long)R + C - N + 1);         }         result += (first + dist);     }       printf("%lld\n", result); }

• C# 6.0

using System;   namespace Csharp {     class Program     {         static void Main(string[] args)         {             long N, K;             string[] inputs = Console.ReadLine().Split();             N = Convert.ToInt64(inputs[0]);             K = Convert.ToInt64(inputs[1]);               long R = 0L, C = 0L, result = 1L;             string moves = Console.ReadLine();             for (int i = 0; i < K; i++)             {                 switch (moves[i])                 {                     case 'U': --R; break;                     case 'D': ++R; break;                     case 'L': --C; break;                     case 'R': ++C; break;                 }                   long lineNo = R + C;                 long first, dist;                 if (lineNo < N)                 {                     first = 1L + Sigma(lineNo);                     dist = (lineNo % 2 == 1) ? R : C;                 }                 else                 {                     first = Sigma(N) + Sigma(N - 1) + Sigma(2L * N - lineNo + 1);                     dist = ((lineNo % 2 == 1) ? R : C) - (lineNo - N);                  }                 result += (first + dist);             }               Console.WriteLine(result);         }                  static long Sigma(long n)         {             return n * (n + 1L) / 2L;         }     } }

• Python 3

from sys import stdin input = stdin.readline   N, K = map(int, input().split()) R, C = 1, 1 firstNumbers = [ 0 ] * (N * 2) firstNumbers[1] = 1   def getLength(lineNo):     if lineNo <= N:         return lineNo          return N - lineNo % N   for i in range(2, N * 2):     firstNumbers[i] = firstNumbers[i - 1] + getLength(i - 1)   def getNumber(r, c):     lineNumber = r + c - 1     firstNumber = firstNumbers[lineNumber]          if lineNumber <= N:         if lineNumber % 2 == 0:             return firstNumber + (r - 1)         else:             return firstNumber + (c - 1)     else:         if lineNumber % 2 == 0:             return firstNumber + (N - c)         else:             return firstNumber + (N - r)      result = 1 for ch in input().rstrip():     if ch == 'U':         R -= 1     elif ch == 'D':         R += 1     elif ch == 'L':         C -= 1     elif ch == 'R':         C += 1       result += getNumber(R, C)   print(result)