CF 102697002 - Triple Product

CF 102697002 - Triple Product

Rating: -
Tags: -
Solve time: 1m 3s
Verified: yes

Solution

Problem Understanding

The task is to compute the value produced by taking two given positive integers and multiplying them together, then multiplying the result by three. The input contains the two factors separately, and the output is the single integer representing this triple product.

The values of the two numbers are small, with each one at most 2000. This means even a direct arithmetic operation is more than sufficient. There is no need for advanced data structures or algorithms because the entire computation is constant time. The main concern is simply performing the multiplication correctly and printing the result.

A few edge cases can cause mistakes in a careless implementation. If both numbers are the smallest possible value, the program must still handle the multiplication normally.

For example:

Input:
1
1

Output:
3

A solution that forgets the final multiplication by three would incorrectly print 1.

Another common mistake is changing the order of operations incorrectly. The expression should be a * b * 3, which is equivalent to (a * b) * 3.

Input:
2000
2000

Output:
12000000

The answer does not fit into a small integer type in some languages, so implementations should use an integer type capable of holding the product. Python integers handle this automatically.

Approaches

The brute-force interpretation of this problem is to search for some more complicated relationship between the numbers, but there is no hidden structure to discover. The required value is directly defined by the arithmetic expression. A brute-force approach that tried different combinations or simulated multiplication would only add unnecessary work while still producing the same result.

The key observation is that the problem asks for one deterministic calculation. Once the two input values are read, the answer is fully determined. The entire solution reduces to reading two integers, multiplying them, multiplying by three, and printing the result.

The brute-force approach is unnecessary, while the direct arithmetic solution finishes in constant time.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n) or worse depending on simulation	O(1)	Too slow and unnecessary
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

1. Read the two integers from the input. Each value represents one factor of the product.
2. Multiply the two values together. This gives the normal product of the two numbers.
3. Multiply the result by three and print it. The extra multiplication is the entire requirement of the problem.

Why it works:

The algorithm follows the mathematical definition of the required output exactly. Since there is only one possible answer for every pair of input values, directly evaluating the expression cannot miss any cases.

Python Solution

import sys

input = sys.stdin.readline

def solve():
    a = int(input())
    b = int(input())
    print(a * b * 3)

if __name__ == "__main__":
    solve()

The program reads the two lines as integers and stores them in a and b. The multiplication is performed in the same order as the formula from the problem, which avoids accidentally omitting the factor of three.

Python's integer implementation supports arbitrary precision, so there is no overflow concern even when the inputs are at their maximum values.

The solution does not need arrays, loops, or additional memory because the input contains only two numbers.

Worked Examples

Example 1

Input:

3
5

Step	a	b	Product
Read input	3	5
Multiply values	3	5	15
Multiply by 3	3	5	45

The calculation confirms that the program performs the required extra multiplication after finding the normal product.

Example 2

Input:

2000
2000

Step	a	b	Product
Read input	2000	2000
Multiply values	2000	2000	4000000
Multiply by 3	2000	2000	12000000

This example checks the largest allowed values and confirms that the arithmetic is still handled correctly.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only two numbers are read and a fixed number of arithmetic operations are performed.
Space	O(1)	The program stores only the two input values.

The constant time and memory usage are far below the limits, so the solution easily satisfies the requirements.

Test Cases

import sys
import io

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    a = int(sys.stdin.readline())
    b = int(sys.stdin.readline())
    print(a * b * 3)

    result = sys.stdout.getvalue()

    sys.stdin = old_stdin
    sys.stdout = old_stdout

    return result

# provided sample
assert run("3\n5\n") == "45\n", "sample 1"

# minimum values
assert run("1\n1\n") == "3\n", "minimum values"

# maximum values
assert run("2000\n2000\n") == "12000000\n", "maximum values"

# uneven factors
assert run("7\n11\n") == "231\n", "normal multiplication"

# one factor equal to one
assert run("1\n2000\n") == "6000\n", "boundary factor"

Test input	Expected output	What it validates
3\n5\n	45	Provided sample behavior
1\n1\n	3	Minimum boundary values
2000\n2000\n	12000000	Maximum multiplication result
7\n11\n	231	General arithmetic correctness
1\n2000\n	6000	Handling a factor equal to one

Edge Cases

For the smallest possible input:

1
1

the algorithm reads both values, computes 1 * 1 * 3, and prints 3. A solution that only multiplies the two inputs would fail because it would ignore the required factor of three.

For the largest possible input:

2000
2000

the algorithm computes 2000 * 2000 * 3 = 12000000. The direct calculation avoids overflow problems in Python and produces the correct result without any special handling.

For cases where one number is much smaller than the other:

1
2000

the algorithm still applies the same formula and outputs 6000. No special cases are needed because multiplication by one naturally preserves the other factor.