CF 102697001 - Square The Number

Rating: -
Tags: -
Solve time: 2m 8s
Verified: yes

Solution

Problem Understanding

The task is to read one positive integer and print its square, meaning the result of multiplying the number by itself. The input contains a single value representing the number to transform, and the output is that same value after applying the squaring operation.

The constraint information is minimal for this problem, but the intended operation is constant time. Even if the number is large, the solution should avoid unnecessary loops or repeated arithmetic. Reading the value, multiplying it once, and printing the result uses a fixed number of operations, so it remains efficient for any normal integer range supported by the language.

The main edge cases come from handling the smallest values and the representation of the number correctly. A common mistake is to assume the number has multiple digits or to perform string-based manipulation instead of arithmetic. For example, the input 1 should produce 1, because 1 * 1 = 1. A careless implementation that expects a two-digit number could fail here.

Another case is a number ending in zero. For input 10, the correct output is 100. A solution that only squares the non-zero part and forgets the place value would produce an incorrect result.

Approaches

The brute-force approach is not really a search, but a repeated multiplication process. A programmer might try to simulate squaring by adding the number to itself many times, performing n additions to compute n * n. This is mathematically correct because multiplication is repeated addition, but it requires n operations. If the input number is large, this immediately becomes impractical.

The useful observation is that the problem asks for a single multiplication. Modern programming languages already provide integer multiplication, so there is no need to reconstruct the operation manually. The entire task reduces to storing the input value and evaluating n * n.

The brute-force method works because repeated addition eventually creates the square, but it fails because it ignores the direct arithmetic operation available. The observation that squaring is simply multiplication of a value by itself reduces the solution to constant time.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n)	O(1)	Too slow for large values
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the integer from input and store it in a variable. The value itself is all the information needed because there are no additional structures or relationships to maintain.
Multiply the value by itself. This directly applies the definition of squaring and avoids unnecessary simulation.
Print the resulting value.

Why it works: The invariant is that the stored variable always represents the original input number. The only transformation performed is multiplication of this value by itself, which is exactly the mathematical definition of a square. Since the required output is precisely that product, the algorithm cannot produce a different result.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
print(n * n)

The program reads the input line and converts it into an integer so arithmetic can be performed directly. The multiplication expression computes the square without any intermediate steps.

The final print outputs the computed value. Python integers can grow beyond fixed machine-size limits, which avoids overflow issues that can appear in some other languages when the input range becomes large.

Worked Examples

For the input 6, the algorithm keeps the value 6, multiplies it by itself, and prints 36.

Input value	Squared value	Output
6	6 × 6 = 36	36

This example demonstrates the normal case where the number has more than one digit and the multiplication produces a larger result.

For the input 10, the algorithm handles the trailing zero naturally.

Input value	Squared value	Output
10	10 × 10 = 100	100

This confirms that the solution does not need special handling for place values or zero digits.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	The algorithm performs one multiplication and one output operation.
Space	O(1)	Only the input value and result are stored.

The solution fits easily within the limits because it does not depend on the size of the input through loops or additional memory.

Test Cases

import sys
import io

def solve(data: str) -> str:
    old_stdin = sys.stdin
    sys.stdin = io.StringIO(data)
    input = sys.stdin.readline

    n = int(input())
    ans = str(n * n)

    sys.stdin = old_stdin
    return ans

# provided sample
assert solve("6\n") == "36", "sample 1"

# minimum-size value
assert solve("1\n") == "1", "minimum value"

# trailing zero case
assert solve("10\n") == "100", "trailing zero"

# larger value
assert solve("100000\n") == "10000000000", "large square"

# repeated digit value
assert solve("111\n") == "12321", "normal multiplication"

Test input	Expected output	What it validates
`1`	`1`	Smallest positive input handling
`10`	`100`	Correct handling of zeros
`100000`	`10000000000`	Large multiplication result
`111`	`12321`	General arithmetic correctness

Edge Cases

For input 1, the algorithm reads 1, computes 1 * 1, and returns 1. This catches implementations that incorrectly assume the result must have more digits than the input.

For input 10, the algorithm computes 10 * 10 and returns 100. A solution based on string tricks or incomplete digit handling could incorrectly return a value without the required zero.

For any large input value, the algorithm still performs the same single multiplication. The execution path does not change based on the magnitude of the number, which is why it remains within the required limits.