There is a tape s with n characters written in a row.
Process q commands in the given order.
L k takes the first character of the tape and attaches it to the end, k times.
R k takes the last character of the tape and attaches it to the front, k times.
F reverses the whole tape (the last character comes first).
W p c overwrites the p-th character from the front of the current tape with the character c.
Q p prints the p-th character from the front of the current tape on one line.
The number p counts the first character as 1. k can be larger than n.
The first line gives s, the second line gives q, and each of the next q lines gives one command. The letters and numbers inside a command are separated by single spaces.
For each Q command print one character on its own line, and on the last line print the whole tape after all commands.
Input
s consists of lowercase English letters and digits, and its length n satisfies 1 ≤ n ≤ 10,000. 1 ≤ q ≤ 10,000. For L and R, 1 ≤ k ≤ 1,000,000,000 (given as plain digits without commas). For W and Q, 1 ≤ p ≤ n. The c of W is one lowercase English letter or digit. There may be no Q command
Output
One character per line for each Q command, then the tape of length n on the last line
Example 1 · stdin
abcdef
10
Q 1
L 2
Q 1
F
Q 1
L 1
Q 1
W 2 x
R 9
Q 6
stdout
a
c
b
a
e
dcbaxe
main.py
Error List00F8 ↓ · ⇧F8 ↑
Input (stdin)
When you run, these lines are fed to input()/stdin in order (the program doesn't stop to ask).
$ Ready
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In other languagesShow solutions by language · 8 languages
Reading by index is s[i] in Python, JS, TS, C#, C, C++ and Craykon, and s.charAt(i) in Java. Writing to that position differs by language. Python s[i] = c raises TypeError, Java has no such syntax, C# and TS give a compile error (read-only), and Craykon stops with 'a string cannot be changed one character at a time'. JS is the most dangerous: s[i] = c is silently ignored (a TypeError only in strict mode), so the tape stays the same. So convert to a list of characters you can change: Python list(s), JS and TS [...s], Java toCharArray(), C# ToCharArray(), and in Craykon Add each character to a list. A C char array and a C++ std::string accept s[i] = c directly. Instead of really rotating or reversing the tape for every command, keep just two numbers meaning 'character i of the current tape is character (start + direction × i) of the original string'; then each command is a single calculation. Trap 1: that index can be negative. Python's % always gives 0 to n − 1, but in C, C++, Java, C#, JS, TS and Craykon -2 % 5 is -2. In JS and TS tape[-2] is undefined without any error, so that character goes missing (join) or "undefined" is printed (when joined with +); Java, C# and Craykon stop with an error; C and C++ read outside the array. Use ((x % n) + n) % n or Java Math.floorMod(x, n). Trap 2: after a flip, 'send the first character to the back' moves the other way in the original string, so the k of L and R must be multiplied by the direction. Trap 3: a flip must not only change the direction; the start must also move to what is currently the last character. Trap 4: the p-th character counted by people is index p - 1 counting from 0. Trap 5: k goes up to a billion, so you cannot rotate one step k times, and adding it unreduced into an int passes 2.1 billion on the third command and wraps around. Reduce it with k % n first.
Pythonmain.py · 31 lines
import syslines = sys.stdin.read().split("\n")tape = list(lines[0].strip()) # 문자열은 한 글자씩 고칠 수 없어서 글자 목록으로 바꾼다n = len(tape)q = int(lines[1])# 띠를 실제로 돌리지 않는다. 지금 띠의 i번 글자(0부터)는 tape[(start + step * i) % n]start, step = 0, 1def where(i): return (start + step * i) % n # Python의 %는 음수에서도 0 ~ n-1을 준다out = []for line in lines[2:2 + q]: p = line.split() if p[0] == "L": start = where(int(p[1]) % n) elif p[0] == "R": start = where(-(int(p[1]) % n)) elif p[0] == "F": start = where(n - 1) # 맨 끝 글자가 새 맨 앞 step = -step elif p[0] == "W": tape[where(int(p[1]) - 1)] = p[2] else: out.append(tape[where(int(p[1]) - 1)])out.append("".join(tape[where(i)] for i in range(n)))print("\n".join(out))
JavaScriptmain.js · 28 lines
const lines = require("fs").readFileSync(0, "utf8").split("\n");const tape = [...lines[0].trim()]; // 문자열은 한 글자씩 고칠 수 없어서 글자 배열로 바꾼다const n = tape.length;const q = Number(lines[1]);// 띠를 실제로 돌리지 않는다. 지금 띠의 i번 글자(0부터)는 tape[(start + step * i) % n]let start = 0, step = 1;// JS의 %는 음수에서 음수를 주므로 n을 더해 한 번 더 나눈다const where = (i) => (((start + step * i) % n) + n) % n;const out = [];for (const line of lines.slice(2, 2 + q)) { const p = line.trim().split(" "); if (p[0] === "L") { start = where(Number(p[1]) % n); } else if (p[0] === "R") { start = where(-(Number(p[1]) % n)); } else if (p[0] === "F") { start = where(n - 1); // 맨 끝 글자가 새 맨 앞 step = -step; } else if (p[0] === "W") { tape[where(Number(p[1]) - 1)] = p[2]; } else { out.push(tape[where(Number(p[1]) - 1)]); }}out.push(tape.map((_, i) => tape[where(i)]).join(""));console.log(out.join("\n"));
TypeScriptmain.ts · 30 lines
import { readFileSync } from "node:fs";const lines: string[] = readFileSync(0, "utf8").split("\n");const tape: string[] = [...lines[0].trim()]; // 문자열은 한 글자씩 고칠 수 없어서 글자 배열로 바꾼다const n = tape.length;const q = Number(lines[1]);// 띠를 실제로 돌리지 않는다. 지금 띠의 i번 글자(0부터)는 tape[(start + step * i) % n]let start = 0, step = 1;// TS의 %는 음수에서 음수를 주므로 n을 더해 한 번 더 나눈다const where = (i: number): number => (((start + step * i) % n) + n) % n;const out: string[] = [];for (const line of lines.slice(2, 2 + q)) { const p = line.trim().split(" "); if (p[0] === "L") { start = where(Number(p[1]) % n); } else if (p[0] === "R") { start = where(-(Number(p[1]) % n)); } else if (p[0] === "F") { start = where(n - 1); // 맨 끝 글자가 새 맨 앞 step = -step; } else if (p[0] === "W") { tape[where(Number(p[1]) - 1)] = p[2]; } else { out.push(tape[where(Number(p[1]) - 1)]); }}out.push(tape.map((_, i) => tape[where(i)]).join(""));console.log(out.join("\n"));
JavaMain.java · 35 lines
import java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;public class Main { static int n, start = 0, step = 1; // 띠를 실제로 돌리지 않는다. 지금 띠의 i번 글자(0부터)가 있는 원래 자리 static int where(int i) { return Math.floorMod(start + step * i, n); // %와 달리 음수에서도 0 ~ n-1을 준다 } public static void main(String[] args) throws IOException { BufferedReader in = new BufferedReader(new InputStreamReader(System.in)); char[] tape = in.readLine().trim().toCharArray(); // String은 고칠 수 없어서 char[]로 바꾼다 n = tape.length; int q = Integer.parseInt(in.readLine().trim()); StringBuilder out = new StringBuilder(); for (int t = 0; t < q; t++) { String[] p = in.readLine().trim().split(" "); switch (p[0]) { case "L" -> start = where(Integer.parseInt(p[1]) % n); case "R" -> start = where(-(Integer.parseInt(p[1]) % n)); case "F" -> { start = where(n - 1); // 맨 끝 글자가 새 맨 앞 step = -step; } case "W" -> tape[where(Integer.parseInt(p[1]) - 1)] = p[2].charAt(0); default -> out.append(tape[where(Integer.parseInt(p[1]) - 1)]).append('\n'); } } for (int i = 0; i < n; i++) out.append(tape[where(i)]); System.out.println(out); }}
C#Main.cs · 46 lines
using System;using System.Text;class Program{ static int n, start = 0, step = 1; // 띠를 실제로 돌리지 않는다. 지금 띠의 i번 글자(0부터)가 있는 원래 자리 static int Where(int i) { return ((start + step * i) % n + n) % n; // C#의 %는 음수에서 음수를 준다 } static void Main() { char[] tape = Console.ReadLine().Trim().ToCharArray(); // string은 고칠 수 없어서 char[]로 바꾼다 n = tape.Length; int q = int.Parse(Console.ReadLine().Trim()); var output = new StringBuilder(); for (int t = 0; t < q; t++) { string[] p = Console.ReadLine().Trim().Split(' '); switch (p[0]) { case "L": start = Where(int.Parse(p[1]) % n); break; case "R": start = Where(-(int.Parse(p[1]) % n)); break; case "F": start = Where(n - 1); // 맨 끝 글자가 새 맨 앞 step = -step; break; case "W": tape[Where(int.Parse(p[1]) - 1)] = p[2][0]; break; default: output.Append(tape[Where(int.Parse(p[1]) - 1)]).Append('\n'); break; } } for (int i = 0; i < n; i++) output.Append(tape[Where(i)]); Console.WriteLine(output); }}
C++main.cpp · 41 lines
#include <iostream>#include <string>int main() { std::ios::sync_with_stdio(false); std::string tape; // std::string은 tape[i] = c로 한 글자씩 고칠 수 있다 int q; std::cin >> tape >> q; const int n = static_cast<int>(tape.size()); // 띠를 실제로 돌리지 않는다. 지금 띠의 i번 글자(0부터)는 tape[(start + step * i) % n] int start = 0, step = 1; // C++의 %는 음수에서 음수를 주므로 n을 더해 한 번 더 나눈다 auto where = [&](int i) { return ((start + step * i) % n + n) % n; }; std::string out; for (int t = 0; t < q; t++) { char cmd, c; int v; std::cin >> cmd; if (cmd == 'L') { std::cin >> v; start = where(v % n); } else if (cmd == 'R') { std::cin >> v; start = where(-(v % n)); } else if (cmd == 'F') { start = where(n - 1); // 맨 끝 글자가 새 맨 앞 step = -step; } else if (cmd == 'W') { std::cin >> v >> c; tape[where(v - 1)] = c; } else { std::cin >> v; out += tape[where(v - 1)]; out += '\n'; } } for (int i = 0; i < n; i++) out += tape[where(i)]; std::cout << out << '\n';}
Cmain.c · 45 lines
#include <stdio.h>#include <string.h>static char tape[10001]; /* char 배열은 한 글자씩 고칠 수 있다 */static int n, start = 0, step = 1;/* 띠를 실제로 돌리지 않는다. 지금 띠의 i번 글자(0부터)가 있는 원래 자리 */static int where(int i) { return ((start + step * i) % n + n) % n; /* C의 %는 음수에서 음수를 준다 */}int main(void) { int q; scanf("%10000s %d", tape, &q); n = (int)strlen(tape); for (int t = 0; t < q; t++) { char cmd, c; int v; scanf(" %c", &cmd); switch (cmd) { case 'L': scanf("%d", &v); start = where(v % n); break; case 'R': scanf("%d", &v); start = where(-(v % n)); break; case 'F': start = where(n - 1); /* 맨 끝 글자가 새 맨 앞 */ step = -step; break; case 'W': scanf("%d %c", &v, &c); tape[where(v - 1)] = c; break; default: scanf("%d", &v); printf("%c\n", tape[where(v - 1)]); } } for (int i = 0; i < n; i++) putchar(tape[where(i)]); putchar('\n'); return 0;}
Craykonmain.cray · 50 lines
줄 = Console.ReadLine().Trim();// 문자열은 한 글자씩 고칠 수 없어서 글자 목록으로 바꾼다tape = [];foreach (c in 줄){ tape.Add(c);}n = tape.Count;q = Console.ReadLine().ToNumber();// 띠를 실제로 돌리지 않는다. 지금 띠의 i번 글자(0부터)는 tape[(start + step * i) % n]start = 0;step = 1;// 크레이콘의 %는 음수에서 음수를 주므로 n을 더해 한 번 더 나눈다자리(i){ return ((start + step * i) % n + n) % n;}for (t = 0; t < q; t++){ p = Console.ReadLine().Trim().Split(" "); switch (p[0]) { case "L": start = 자리(p[1].ToNumber() % n); break; case "R": start = 자리(-(p[1].ToNumber() % n)); break; case "F": // 맨 끝 글자가 새 맨 앞 start = 자리(n - 1); step = -step; break; case "W": tape[자리(p[1].ToNumber() - 1)] = p[2]; break; default: Console.WriteLine(tape[자리(p[1].ToNumber() - 1)]); break; }}끝 = [];for (i = 0; i < n; i++){ 끝.Add(tape[자리(i)]);}Console.WriteLine(끝.Join(""));