Fill a board of r rows and c columns with the numbers 1 to r × c in a spiral.
The first line gives r and c.
Write 1 in the top-left cell and keep writing the next number while moving right.
If the next cell is outside the board or already has a number, turn 90 degrees clockwise (right → down → left → up → right) and write in the cell that way.
When the board is full, print it row by row.
Right-align each number to a width equal to the number of digits of r × c (pad the front with spaces), with one space between numbers.
Input
1 ≤ r ≤ 50, 1 ≤ c ≤ 50
Output
r lines, c numbers per line
Example 1 · stdin
3 4
stdout
1 2 3 4
10 11 12 5
9 8 7 6
Example 2 · stdin
4 3
stdout
1 2 3
10 11 4
9 12 5
8 7 6
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
Open this on a computer to write and run code. On mobile you can read the problem and the solutions.
In other languagesShow solutions by language · 8 languages
The first trap is how you build a board filled with 0. Python [[0] * c] * r puts one row list in r times, so writing one cell changes that cell in every row. Build each row anew with [[0] * c for _ in range(r)]. JS/TS new Array(r).fill(new Array(c).fill(0)) also puts the same array in every slot, so write Array.from({ length: r }, () => new Array(c).fill(0)). Java new int[r][c], C# new int[r, c] and a global C array start at 0, and C++ vector(r, vector<int>(c, 0)) copies each row. Craykon NewList(r, NewList(c, 0)) also copies the value into each slot. Turning clockwise means changing (dy, dx) into (dx, -dy), which is a swap of two values. Python dy, dx = dx, -dy, JS/TS [dy, dx] = [dx, -dy], C# (dy, dx) = (dx, -dy) change both at once, and C++ can write std::swap(dy, dx); dx = -dx; without a temporary, but Java, C and Craykon need a temporary t = dy; dy = dx; dx = -t;. Writing dy = dx; dx = -dy; without the temporary uses the already changed dy on the second line and walks diagonally. The width is the number of digits of r × c (3 for 100 cells, for example). Align with Python str(v).rjust(w), JS/TS padStart(w), Java String.format("%" + w + "d", v), C printf("%*d", w, v), C++ std::setw(w), C# PadLeft(w), Craykon PadLeft(w).
Pythonmain.py · 14 lines
r, c = map(int, input().split())grid = [[0] * c for _ in range(r)] # 줄마다 새 목록(같은 목록을 r번 넣지 않게)y, x = 0, 0dy, dx = 0, 1 # 오른쪽으로 시작for k in range(1, r * c + 1): grid[y][x] = k ny, nx = y + dy, x + dx if not (0 <= ny < r and 0 <= nx < c and grid[ny][nx] == 0): dy, dx = dx, -dy # 시계 방향으로 90도: 두 값을 맞바꾸며 부호 하나를 뒤집는다 ny, nx = y + dy, x + dx y, x = ny, nxwidth = len(str(r * c))for row in grid: print(" ".join(str(v).rjust(width) for v in row))
JavaScriptmain.js · 18 lines
const [r, c] = require("fs").readFileSync(0, "utf8").trim().split(/\s+/).map(Number);// 줄마다 새 배열(fill에 배열 하나를 주면 모든 줄이 같은 배열이 된다)const grid = Array.from({ length: r }, () => new Array(c).fill(0));let y = 0, x = 0;let dy = 0, dx = 1; // 오른쪽으로 시작for (let k = 1; k <= r * c; k++) { grid[y][x] = k; let ny = y + dy, nx = x + dx; if (!(ny >= 0 && ny < r && nx >= 0 && nx < c && grid[ny][nx] === 0)) { [dy, dx] = [dx, -dy]; // 시계 방향으로 90도 ny = y + dy; nx = x + dx; } y = ny; x = nx;}const width = String(r * c).length;console.log(grid.map((row) => row.map((v) => String(v).padStart(width)).join(" ")).join("\n"));
TypeScriptmain.ts · 19 lines
import { readFileSync } from "node:fs";const [r, c]: number[] = readFileSync(0, "utf8").trim().split(/\s+/).map(Number);// 줄마다 새 배열(fill에 배열 하나를 주면 모든 줄이 같은 배열이 된다)const grid: number[][] = Array.from({ length: r }, () => new Array<number>(c).fill(0));let y = 0, x = 0;let dy = 0, dx = 1; // 오른쪽으로 시작for (let k = 1; k <= r * c; k++) { grid[y][x] = k; let ny = y + dy, nx = x + dx; if (!(ny >= 0 && ny < r && nx >= 0 && nx < c && grid[ny][nx] === 0)) { [dy, dx] = [dx, -dy]; // 시계 방향으로 90도 ny = y + dy; nx = x + dx; } y = ny; x = nx;}const width: number = String(r * c).length;console.log(grid.map((row) => row.map((v) => String(v).padStart(width)).join(" ")).join("\n"));
JavaMain.java · 35 lines
import java.util.Scanner;public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int r = sc.nextInt(); int c = sc.nextInt(); int[][] grid = new int[r][c]; // 칸마다 0으로 채워져 있다 int y = 0, x = 0; int dy = 0, dx = 1; // 오른쪽으로 시작 for (int k = 1; k <= r * c; k++) { grid[y][x] = k; int ny = y + dy, nx = x + dx; if (!(ny >= 0 && ny < r && nx >= 0 && nx < c && grid[ny][nx] == 0)) { int t = dy; // 시계 방향으로 90도: 임시 변수로 맞바꾼다 dy = dx; dx = -t; ny = y + dy; nx = x + dx; } y = ny; x = nx; } String fmt = "%" + String.valueOf(r * c).length() + "d"; StringBuilder out = new StringBuilder(); for (int[] row : grid) { for (int j = 0; j < c; j++) { if (j > 0) out.append(' '); out.append(String.format(fmt, row[j])); } out.append('\n'); } System.out.print(out); }}
C#Main.cs · 40 lines
using System;using System.Linq;using System.Text;class Program{ static void Main() { int[] rc = Console.ReadLine().Split(' ', StringSplitOptions.RemoveEmptyEntries).Select(int.Parse).ToArray(); int r = rc[0], c = rc[1]; int[,] grid = new int[r, c]; // 칸마다 0으로 채워져 있다 int y = 0, x = 0; int dy = 0, dx = 1; // 오른쪽으로 시작 for (int k = 1; k <= r * c; k++) { grid[y, x] = k; int ny = y + dy, nx = x + dx; if (!(ny >= 0 && ny < r && nx >= 0 && nx < c && grid[ny, nx] == 0)) { (dy, dx) = (dx, -dy); // 시계 방향으로 90도: 튜플로 맞바꾼다 ny = y + dy; nx = x + dx; } y = ny; x = nx; } int width = (r * c).ToString().Length; var sb = new StringBuilder(); for (int i = 0; i < r; i++) { for (int j = 0; j < c; j++) { if (j > 0) sb.Append(' '); sb.Append(grid[i, j].ToString().PadLeft(width)); } sb.Append('\n'); } Console.Write(sb); }}
C++main.cpp · 33 lines
#include <iomanip>#include <iostream>#include <string>#include <utility>#include <vector>int main() { int r, c; std::cin >> r >> c; std::vector<std::vector<int>> grid(r, std::vector<int>(c, 0)); // 줄마다 따로 복사된다 int y = 0, x = 0; int dy = 0, dx = 1; // 오른쪽으로 시작 for (int k = 1; k <= r * c; k++) { grid[y][x] = k; int ny = y + dy, nx = x + dx; if (!(ny >= 0 && ny < r && nx >= 0 && nx < c && grid[ny][nx] == 0)) { std::swap(dy, dx); // 시계 방향으로 90도: 맞바꾸고 dx = -dx; // 부호 하나를 뒤집는다 ny = y + dy; nx = x + dx; } y = ny; x = nx; } int width = std::to_string(r * c).size(); for (const auto& row : grid) { for (int j = 0; j < c; j++) { if (j > 0) std::cout << ' '; std::cout << std::setw(width) << row[j]; } std::cout << '\n'; }}
Cmain.c · 29 lines
#include <stdio.h>int grid[50][50]; // 전역 배열은 0으로 채워져 있다int main(void) { int r, c; scanf("%d %d", &r, &c); int y = 0, x = 0; int dy = 0, dx = 1; // 오른쪽으로 시작 for (int k = 1; k <= r * c; k++) { grid[y][x] = k; int ny = y + dy, nx = x + dx; if (!(ny >= 0 && ny < r && nx >= 0 && nx < c && grid[ny][nx] == 0)) { int t = dy; // 시계 방향으로 90도: 임시 변수로 맞바꾼다 dy = dx; dx = -t; ny = y + dy; nx = x + dx; } y = ny; x = nx; } int width = snprintf(NULL, 0, "%d", r * c); // r*c의 자릿수 for (int i = 0; i < r; i++) { for (int j = 0; j < c; j++) printf(j ? " %*d" : "%*d", width, grid[i][j]); putchar('\n'); } return 0;}
Craykonmain.cray · 35 lines
rc = Console.ReadNumbers();r = rc[0];c = rc[1];// 새목록에 목록을 주면 줄마다 따로 복사해 넣는다grid = NewList(r, NewList(c, 0));y = 0;x = 0;dy = 0;dx = 1; // 오른쪽으로 시작for (k = 1; k <= r * c; k++){ grid[y][x] = k; ny = y + dy; nx = x + dx; if (!(ny >= 0 && ny < r && nx >= 0 && nx < c && grid[ny][nx] == 0)) { t = dy; // 시계 방향으로 90도: 임시 변수로 맞바꾼다 dy = dx; dx = -t; ny = y + dy; nx = x + dx; } y = ny; x = nx;}width = ("" + r * c).Length;foreach (row in grid){ cells = []; foreach (v in row) { cells.Add(("" + v).PadLeft(width)); } Console.WriteLine(cells.Join(" "));}