CF 106185A - 2025

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Solve time: 42s
Verified: yes

Solution

Problem Understanding

We are given a sequence of test cases. Each test case contains a single integer $n$, and for each one we conceptually build an $n \times n$ multiplication table where the cell in row $a$ and column $b$ contains the product $a \cdot b$. The task is to compute the sum of all entries in this table.

So for a fixed $n$, the required value is

$$\sum_{a=1}^{n} \sum_{b=1}^{n} a \cdot b.$$

The input may contain up to 100 values of $n$, each between 1 and 100 inclusive, and input ends when a single zero appears.

The bounds are extremely small. Even the most naive double loop over $a$ and $b$ is at most $100 \times 100 = 10^4$ operations per test case, which across 100 test cases is trivial. This already tells us that any reasonable $O(n^2)$ or even $O(n^3)$ solution would pass, but the structure of the expression allows something cleaner.

Edge cases are mostly about small $n$. When $n = 1$, the answer is $1$. When $n = 2$, the table is

$$\begin{matrix} 1 & 2 \ 2 & 4 \end{matrix}$$

and the sum is $9$. A wrong approach typically appears when one tries to interpret the problem as summing only diagonals or only distinct products, but the problem explicitly counts every cell including duplicates like $a \cdot b$ and $b \cdot a$ as separate entries.

Another subtle failure mode is treating the sum as involving unique pairs $a \le b$, which would halve off-diagonal contributions incorrectly.

Approaches

The most direct approach is to simulate the table. For each test case, iterate over all pairs $(a, b)$, compute the product, and accumulate it. This is straightforward and correct because it mirrors the definition exactly. The cost per test case is $O(n^2)$, and with $n \le 100$, this means at most $10^4$ multiplications per case, which is negligible.

However, the expression has strong separability. The sum

$$\sum_{a=1}^{n} \sum_{b=1}^{n} a \cdot b$$

can be rearranged by factoring:

$$\left(\sum_{a=1}^{n} a\right) \cdot \left(\sum_{b=1}^{n} b\right).$$

Both sums are identical, so the result becomes:

$$\left(\sum_{i=1}^{n} i\right)^2.$$

The inner sum is the standard arithmetic series:

$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}.$$

So the final answer is:

$$\left(\frac{n(n+1)}{2}\right)^2.$$

The key observation is recognizing that the double sum over a product splits into a product of sums because the indices are independent. This reduces each test case to constant time arithmetic.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	$O(n^2)$	$O(1)$	Accepted
Optimal	$O(1)$	$O(1)$	Accepted

Algorithm Walkthrough

We process test cases one by one until we encounter a zero.

Read an integer $n$. If $n = 0$, stop processing. This is necessary because zero is a sentinel and not part of the input domain.
Compute the sum of integers from 1 to $n$ using the formula $s = n(n+1)/2$. This replaces iterating through all values and is valid because the sequence is arithmetic with constant difference 1.
Square the value $s$ to obtain the final answer. This corresponds to combining two independent identical sums over rows and columns of the multiplication table.
Output the result immediately.

Why it works: each cell $(a, b)$ contributes $a \cdot b$. Since $a$ appears independently of $b$, every choice of $a$ pairs with every choice of $b$, so the total contribution of each dimension multiplies. This creates a complete factorization of the double sum into a product of two identical linear sums, ensuring no interaction terms are missed or overcounted.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    out = []
    while True:
        line = input().strip()
        if not line:
            break
        n = int(line)
        if n == 0:
            break
        s = n * (n + 1) // 2
        out.append(str(s * s))
    sys.stdout.write("\n".join(out))

if __name__ == "__main__":
    solve()

The implementation reads each test case until the terminating zero. For each $n$, it computes the triangular number using integer arithmetic, ensuring no floating-point precision issues, and then squares it. The use of integer division guarantees correctness because $n(n+1)$ is always even.

A subtle point is handling input termination correctly. The loop must stop at zero rather than attempting to process it, since including it would incorrectly produce an extra output line.

Worked Examples

Example 1

Input:

2
3
0

For $n = 2$, we compute $s = 2 \cdot 3 / 2 = 3$, so result is $9$.

For $n = 3$, $s = 3 \cdot 4 / 2 = 6$, so result is $36$.

n	s = n(n+1)/2	s²
2	3	9
3	6	36

This confirms that each test case is independent and computed via the same transformation.

Example 2

Input:

1
5
0

For $n = 1$, $s = 1$, result $1$.

For $n = 5$, $s = 15$, result $225$.

n	s	s²
1	1	1
5	15	225

This shows correctness at the smallest boundary and a larger nontrivial case where many off-diagonal products contribute.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(T)$	Each test case uses constant arithmetic operations
Space	$O(1)$	Only a few integers are stored regardless of input size

The constraints allow up to 100 test cases with $n \le 100$, so even a naive solution would be fast. The optimized form reduces each case to constant time, making runtime effectively instantaneous.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys as _sys
    from math import prod

    input = sys.stdin.readline
    out = []

    while True:
        line = input().strip()
        if not line:
            break
        n = int(line)
        if n == 0:
            break
        s = n * (n + 1) // 2
        out.append(str(s * s))

    return "\n".join(out)

# provided samples (illustrative, since original sample omitted)
assert run("1\n0\n") == "1"
assert run("2\n3\n0\n") == "9\n36"

# custom cases
assert run("1\n") == "1", "minimum input"
assert run("2\n") == "9", "small table correctness"
assert run("10\n") == str((10*11//2)**2), "formula consistency"
assert run("100\n0\n") == str((100*101//2)**2), "maximum n boundary"

Test input	Expected output	What it validates
1	1	smallest possible table
2	9	off-diagonal handling
10	3025²	correctness of formula
100	(5050²)	upper bound stability

Edge Cases

For $n = 1$, the grid has a single cell containing $1 \cdot 1$. The algorithm computes $s = 1$, and outputs $1$, matching the table exactly.

For $n = 2$, the grid contains four products: 1, 2, 2, and 4. The algorithm computes $s = 3$, then squares it to get 9. This matches the sum of all four entries, confirming that symmetric off-diagonal contributions are not double counted or missed.

For larger $n$, such as 100, the computation remains stable because all intermediate values fit comfortably in 64-bit integers. The structure avoids iterative accumulation, so there is no risk of cumulative floating-point error or overflow in intermediate loops.