Tutorial:

We have two competitors, a tortoise and a monkey, each trying to reach position \(N\). The tortoise moves a fixed distance \(S\) in one step, so it needs \(\lceil \tfrac{N}{S} \rceil\) steps to finish. Meanwhile, the monkey can jump a distance between \(1\) and \(X\) per step. In the worst case (if we consider the maximum possible jumps for minimal steps), it effectively needs \(\lceil \tfrac{N}{X} \rceil\) steps to finish.

1. Calculate Tortoise Steps: 
   \[
   \text{tortoise\_steps} = \Bigl\lceil \frac{N}{S} \Bigr\rceil
   \]

2. Calculate Monkey Steps: 
   \[
   \text{monkey\_steps} = \Bigl\lceil \frac{N}{X} \Bigr\rceil
   \]

3. Compare: 
   - If \(\text{tortoise\_steps} < \text{monkey\_steps}\), the tortoise wins.  
   - If \(\text{monkey\_steps} < \text{tortoise\_steps}\), the monkey wins.  
   - Otherwise, it’s a tie.

Finally, output "Tortoise", "Monkey", or "Tie" based on the comparison.

void solve() {
    int N, S, X;
    cin >> N >> S >> X;
    
    int tortoise_steps = ceil((double)N / S);
    int monkey_steps = ceil((double)N / X);
    
    string winner;
    if (tortoise_steps < monkey_steps) {
        winner = "Tortoise";
    } else if (monkey_steps < tortoise_steps) {
        winner = "Monkey";
    } else {
        winner = "Tie";
    }
    
    cout << winner << e;
}

Hints:
1- Compute the total available area by multiplying the number of sheets by the area of each sheet.  
2- Calculate the tray’s area as the square of its side length.  
3- Compare the two areas: if total sheet area is at least the tray’s area, output YES; otherwise, NO.

Solution:  
To solve the problem, note that each Roqaq sheet is a square with area \(b^2\), and with \(n\) boxes containing \(s\) sheets each, the total available area is \(n \times s \times b^2\). The tray is also a square, with area \(a^2\). The cook can cover one full layer on the tray if the available area is at least as large as the tray’s area; that is, if \(n \times s \times b^2 \ge a^2\). A correct implementation should compare these squared values directly to avoid any floating-point precision issues. The provided code mistakenly uses \(\sqrt{n \times s} \times b\) for the comparison, which might work in some cases but is not as robust.

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Tutorial:

In this problem, you are asked to count the number of ways to choose exactly three bulbs to be turned on from a string of \( n \) bulbs, which is simply the combination "n choose 3" (i.e. \(\binom{n}{3}\)). The formula for this is:

\[
\binom{n}{3} = \frac{n \times (n-1) \times (n-2)}{6}
\]

Given that \( n \) can be as large as \( 10^{15} \), directly computing the product \( n \times (n-1) \times (n-2) \) might result in numbers that exceed the limits of built-in data types. To handle this, the solution first simplifies the expression by dividing by 6 before performing multiplications. This is done by taking the numbers \( n \), \( n-1 \), and \( n-2 \) and sequentially canceling common factors with 6 using the greatest common divisor (GCD). After the cancellation, the product is computed in a way that avoids overflow. 

Because the resulting number can be extremely large, the solution uses a custom string multiplication function (mul) to multiply numbers represented as strings, which simulates big integer arithmetic. The final product, which is the answer to \(\binom{n}{3}\), is printed as a string.

Step-by-Step Overview:
1. Input Reading: Read the number of test cases and for each test case, read the integer \( n \).
2. Prepare Terms: Let \( a = n \), \( b = n - 1 \), and \( c = n - 2 \).
3. Simplify the Fraction: Initialize \( x = 6 \) (the denominator). For each of \( a \), \( b \), and \( c \), use the GCD to cancel factors from \( x \) so that division by 6 is handled exactly, which avoids overflow.
4. Multiply with Big Integer Arithmetic: Use a custom string multiplication function to multiply the simplified numbers together. The multiplication is performed on string representations of the numbers to handle the extremely large result.
5. Output: Print the resulting product, which represents the number of ways three bulbs can be turned on.

This method efficiently computes \(\binom{n}{3}\) even for very large \( n \), taking care of precision issues with large numbers by using big integer simulation via strings.

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Tutorial:

We model the orange as a perfect sphere whose diameter is \(n \times x\). Consequently, its radius is \(\tfrac{n \times x}{2}\). When you make a horizontal slice at a distance of \(i \times x\) from the top (where \(i\) is counted in “slice-thickness” units), the cross-section is a circle whose radius \(r\) satisfies:

\[
r^2 \;=\; R^2 \;-\;\bigl(R - i\,x\bigr)^2
\]
where \(R = \tfrac{n\,x}{2}\). Expanding and simplifying yields:
\[
r^2 \;=\; x^2 \,\bigl[\,i\,\bigl(n - i\bigr)\bigr].
\]

Dr. Kawthar “dislikes the first and last slices,” meaning \(i\) must be strictly between \(0\) and \(n\). So we look for an integer \(i\in[1,n-1]\) such that
\[
x^2 \;i\,\bigl(n - i\bigr) \;=\; r^2.
\]

- If such an \(i\) exists, that slice’s top has the desired squared radius \(r^2\).  
- If no integer \(i\) satisfies the equation, we print \(-1\).  

In the code, this check is done either by iterating through possible \(i\) values or by using binary search to find whether any \(i\) in the valid range yields the required \(r^2\). If found, we output \(i+1\) (the slice number), else \(-1\).

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