3533번 - Explicit Formula 다국어

Consider 10 Boolean variables x₁, x₂, x₃, x₄, x₅, x₆, x₇, x₈, x₉, and x₁₀. Consider all pairs and triplets of distinct variables among these ten. (There are 45 pairs and 120 triplets.) Count the number of pairs and triplets that contain at least one variable equal to 1. Set f(x₁, x₂, x₃, x₄, x₅, x₆, x₇, x₈, x₉, x₁₀) = 1 if this number is odd and f(x₁, x₂, x₃, x₄, x₅, x₆, x₇, x₈, x₉, x₁₀) = 0 if this number is even.

Here’s an explicit formula that represents the function f(x₁, x₂, x₃, x₄, x₅, x₆, x₇, x₈, x₉, x₁₀) correctly:

f(x₁, x₂, x₃, x₄, x₅, x₆, x₇, x₈, x₉, x₁₀) = (x₁ ∨ x₂) ⊕ (x₁ ∨ x₃) ⊕ (x₁ ∨ x₄) ⊕ (x₁ ∨ x₅) ⊕ (x₁ ∨ x₆) ⊕ (x₁ ∨ x₇) ⊕ (x₁ ∨ x₈) ⊕ (x₁ ∨ x₉) ⊕ (x₁ ∨ x₁₀) ⊕ (x₂ ∨ x₃) ⊕ (x₂ ∨ x₄) ⊕ (x₂ ∨ x₅) ⊕ (x₂ ∨ x₆) ⊕ (x₂ ∨ x₇) ⊕ (x₂ ∨ x₈) ⊕ (x₂ ∨ x₉) ⊕ (x₂ ∨ x₁₀) ⊕ (x₃ ∨ x₄) ⊕ (x₃ ∨ x₅) ⊕ (x₃ ∨ x₆) ⊕ (x₃ ∨ x₇) ⊕ (x₃ ∨ x₈) ⊕ (x₃ ∨ x₉) ⊕ (x₃ ∨ x₁₀) ⊕ (x₄ ∨ x₅) ⊕ (x₄ ∨ x₆) ⊕ (x₄ ∨ x₇) ⊕ (x₄ ∨ x₈) ⊕ (x₄ ∨ x₉) ⊕ (x₄ ∨ x₁₀) ⊕ (x₅ ∨ x₆) ⊕ (x₅ ∨ x₇) ⊕ (x₅ ∨ x₈) ⊕ (x₅ ∨ x₉) ⊕ (x₅ ∨ x₁₀) ⊕ (x₆ ∨ x₇) ⊕ (x₆ ∨ x₈) ⊕ (x₆ ∨ x₉) ⊕ (x₆ ∨ x₁₀) ⊕ (x₇ ∨ x₈) ⊕ (x₇ ∨ x₉) ⊕ (x₇ ∨ x₁₀) ⊕ (x₈ ∨ x₉) ⊕ (x₈ ∨ x₁₀) ⊕ (x₉ ∨ x₁₀) ⊕ (x₁ ∨ x₂ ∨ x₃) ⊕ (x₁ ∨ x₂ ∨ x₄) ⊕ (x₁ ∨ x₂ ∨ x₅) ⊕ (x₁ ∨ x₂ ∨ x₆) ⊕ (x₁ ∨ x₂ ∨ x₇) ⊕ (x₁ ∨ x₂ ∨ x₈) ⊕ (x₁ ∨ x₂ ∨ x₉) ⊕ (x₁ ∨ x₂ ∨ x₁₀) ⊕ (x₁ ∨ x₃ ∨ x₄) ⊕ (x₁ ∨ x₃ ∨ x₅) ⊕ (x₁ ∨ x₃ ∨ x₆) ⊕ (x₁ ∨ x₃ ∨ x₇) ⊕ (x₁ ∨ x₃ ∨ x₈) ⊕ (x₁ ∨ x₃ ∨ x₉) ⊕ (x₁ ∨ x₃ ∨ x₁₀) ⊕ (x₁ ∨ x₄ ∨ x₅) ⊕ (x₁ ∨ x₄ ∨ x₆) ⊕ (x₁ ∨ x₄ ∨ x₇) ⊕ (x₁ ∨ x₄ ∨ x₈) ⊕ (x₁ ∨ x₄ ∨ x₉) ⊕ (x₁ ∨ x₄ ∨ x₁₀) ⊕ (x₁ ∨ x₅ ∨ x₆) ⊕ (x₁ ∨ x₅ ∨ x₇) ⊕ (x₁ ∨ x₅ ∨ x₈) ⊕ (x₁ ∨ x₅ ∨ x₉) ⊕ (x₁ ∨ x₅ ∨ x₁₀) ⊕ (x₁ ∨ x₆ ∨ x₇) ⊕ (x₁ ∨ x₆ ∨ x₈) ⊕ (x₁ ∨ x₆ ∨ x₉) ⊕ (x₁ ∨ x₆ ∨ x₁₀) ⊕ (x₁ ∨ x₇ ∨ x₈) ⊕ (x₁ ∨ x₇ ∨ x₉) ⊕ (x₁ ∨ x₇ ∨ x₁₀) ⊕ (x₁ ∨ x₈ ∨ x₉) ⊕ (x₁ ∨ x₈ ∨ x₁₀) ⊕ (x₁ ∨ x₉ ∨ x₁₀) ⊕ (x₂ ∨ x₃ ∨ x₄) ⊕ (x₂ ∨ x₃ ∨ x₅) ⊕ (x₂ ∨ x₃ ∨ x₆) ⊕ (x₂ ∨ x₃ ∨ x₇) ⊕ (x₂ ∨ x₃ ∨ x₈) ⊕ (x₂ ∨ x₃ ∨ x₉) ⊕ (x₂ ∨ x₃ ∨ x₁₀) ⊕ (x₂ ∨ x₄ ∨ x₅) ⊕ (x₂ ∨ x₄ ∨ x₆) ⊕ (x₂ ∨ x₄ ∨ x₇) ⊕ (x₂ ∨ x₄ ∨ x₈) ⊕ (x₂ ∨ x₄ ∨ x₉) ⊕ (x₂ ∨ x₄ ∨ x₁₀) ⊕ (x₂ ∨ x₅ ∨ x₆) ⊕ (x₂ ∨ x₅ ∨ x₇) ⊕ (x₂ ∨ x₅ ∨ x₈) ⊕ (x₂ ∨ x₅ ∨ x₉) ⊕ (x₂ ∨ x₅ ∨ x₁₀) ⊕ (x₂ ∨ x₆ ∨ x₇) ⊕ (x₂ ∨ x₆ ∨ x₈) ⊕ (x₂ ∨ x₆ ∨ x₉) ⊕ (x₂ ∨ x₆ ∨ x₁₀) ⊕ (x₂ ∨ x₇ ∨ x₈) ⊕ (x₂ ∨ x₇ ∨ x₉) ⊕ (x₂ ∨ x₇ ∨ x₁₀) ⊕ (x₂ ∨ x₈ ∨ x₉) ⊕ (x₂ ∨ x₈ ∨ x₁₀) ⊕ (x₂ ∨ x₉ ∨ x₁₀) ⊕ (x₃ ∨ x₄ ∨ x₅) ⊕ (x₃ ∨ x₄ ∨ x₆) ⊕ (x₃ ∨ x₄ ∨ x₇) ⊕ (x₃ ∨ x₄ ∨ x₈) ⊕ (x₃ ∨ x₄ ∨ x₉) ⊕ (x₃ ∨ x₄ ∨ x₁₀) ⊕ (x₃ ∨ x₅ ∨ x₆) ⊕ (x₃ ∨ x₅ ∨ x₇) ⊕ (x₃ ∨ x₅ ∨ x₈) ⊕ (x₃ ∨ x₅ ∨ x₉) ⊕ (x₃ ∨ x₅ ∨ x₁₀) ⊕ (x₃ ∨ x₆ ∨ x₇) ⊕ (x₃ ∨ x₆ ∨ x₈) ⊕ (x₃ ∨ x₆ ∨ x₉) ⊕ (x₃ ∨ x₆ ∨ x₁₀) ⊕ (x₃ ∨ x₇ ∨ x₈) ⊕ (x₃ ∨ x₇ ∨ x₉) ⊕ (x₃ ∨ x₇ ∨ x₁₀) ⊕ (x₃ ∨ x₈ ∨ x₉) ⊕ (x₃ ∨ x₈ ∨ x₁₀) ⊕ (x₃ ∨ x₉ ∨ x₁₀) ⊕ (x₄ ∨ x₅ ∨ x₆) ⊕ (x₄ ∨ x₅ ∨ x₇) ⊕ (x₄ ∨ x₅ ∨ x₈) ⊕ (x₄ ∨ x₅ ∨ x₉) ⊕ (x₄ ∨ x₅ ∨ x₁₀) ⊕ (x₄ ∨ x₆ ∨ x₇) ⊕ (x₄ ∨ x₆ ∨ x₈) ⊕ (x₄ ∨ x₆ ∨ x₉) ⊕ (x₄ ∨ x₆ ∨ x₁₀) ⊕ (x₄ ∨ x₇ ∨ x₈) ⊕ (x₄ ∨ x₇ ∨ x₉) ⊕ (x₄ ∨ x₇ ∨ x₁₀) ⊕ (x₄ ∨ x₈ ∨ x₉) ⊕ (x₄ ∨ x₈ ∨ x₁₀) ⊕ (x₄ ∨ x₉ ∨ x₁₀) ⊕ (x₅ ∨ x₆ ∨ x₇) ⊕ (x₅ ∨ x₆ ∨ x₈) ⊕ (x₅ ∨ x₆ ∨ x₉) ⊕ (x₅ ∨ x₆ ∨ x₁₀) ⊕ (x₅ ∨ x₇ ∨ x₈) ⊕ (x₅ ∨ x₇ ∨ x₉) ⊕ (x₅ ∨ x₇ ∨ x₁₀) ⊕ (x₅ ∨ x₈ ∨ x₉) ⊕ (x₅ ∨ x₈ ∨ x₁₀) ⊕ (x₅ ∨ x₉ ∨ x₁₀) ⊕ (x₆ ∨ x₇ ∨ x₈) ⊕ (x₆ ∨ x₇ ∨ x₉) ⊕ (x₆ ∨ x₇ ∨ x₁₀) ⊕ (x₆ ∨ x₈ ∨ x₉) ⊕ (x₆ ∨ x₈ ∨ x₁₀) ⊕ (x₆ ∨ x₉ ∨ x₁₀) ⊕ (x₇ ∨ x₈ ∨ x₉) ⊕ (x₇ ∨ x₈ ∨ x₁₀) ⊕ (x₇ ∨ x₉ ∨ x₁₀) ⊕ (x₈ ∨ x₉ ∨ x₁₀)

In this formula ∨ stands for logical or, and ⊕ stands for exclusive or (xor). Remember that in C++ and Java these two binary operators are denoted as “||” and “^”.

Given the values of x₁, x₂, x₃, x₄, x₅, x₆, x₇, x₈, x₉, x₁₀, calculate the value of f(x₁, x₂, . . . , x₁₀).

The input file contains 10 numbers x₁, x₂, x₃, x₄, x₅, x₆, x₇, x₈, x₉, and x₁₀. Each of them is either 0 or 1.

Output a single value — f(x₁, x₂, x₃, x₄, x₅, x₆, x₇, x₈, x₉, x₁₀).