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Doubling Grains on a Chessboard

Calculate a square’s grain count and the total on all 64 squares.

Lo que aprenderás

  • Compute doubling with powers of two
  • Sum a geometric series
  • Work with values near the 64-bit limit

Problema

Ver traducción

Square 1 holds one grain. Each later square holds twice as many grains as the previous square. Return both the count on the requested square and the total on squares 1 through 64. Write each count as decimal text, because the board total is too large for a signed 64-bit integer.

Implement chessboard_grains(square).

Input

square: The requested square, numbered from 1.

Output

Return two decimal strings: [count on the requested square, total on all 64 squares].

Restricciones

1 ≤ square ≤ 64. Output strings contain decimal digits only, with no commas, spaces, or leading zeros.

Ejemplos

Ejemplo 1
Entrada:
square = 6

Salida esperada:
expected = ["32", "18446744073709551615"]
Entrada:
long long square = 6;

Salida esperada:
std::vector<std::string> expected = {"32", "18446744073709551615"};
Entrada:
long square = 6;

Salida esperada:
string[] expected = new string[] {"32", "18446744073709551615"};

Explicación

The first six counts are 1, 2, 4, 8, 16, 32. The board total is 1 + 2 + … + 2⁶³ = 2⁶⁴ − 1.

Ejemplo 2
Entrada:
square = 1

Salida esperada:
expected = ["1", "18446744073709551615"]
Entrada:
long long square = 1;

Salida esperada:
std::vector<std::string> expected = {"1", "18446744073709551615"};
Entrada:
long square = 1;

Salida esperada:
string[] expected = new string[] {"1", "18446744073709551615"};

Explicación

The first square holds 1 grain. The second output always refers to all 64 squares, not just the requested prefix.

Cómo abordarlo

Inténtalo primero por tu cuenta y luego abre las pistas una a una.

Pista 1

Each square doubles the previous square, so the requested count depends only on its position.

Pista 2

The total across all 64 squares is a geometric progression, and it grows far beyond normal integer limits.

Pista 3

Return both values as plain decimal text, and watch that the first square is one, not zero.

Complejidad esperada

O(1) time, O(1) extra space

Referencias

Este reto está adaptado para CodeUtility. Sigue el enunciado y las restricciones indicadas arriba.

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