CF 1505F - Math

CF 1505F - Math

Rating: 2200
Tags: *special, math
Solve time: 5m 35s
Verified: no

Solution

Problem Understanding

The statement is given as a picture, but the task itself is extremely small. We receive a single integer a, whose value lies between -100 and 100. The expression shown in the image evaluates to the same number a, so the answer that must be printed is simply the input value itself.

Since there is only one integer and its magnitude is tiny, any complexity discussion is almost irrelevant. Even a constant amount of work is more than enough. Reading the number and printing it immediately runs in negligible time and uses constant memory.

The only situations where an implementation could go wrong are caused by misunderstanding the picture rather than by algorithmic issues.

Consider the input

1

The correct output is

1

A reader who interprets the image incorrectly and performs some unnecessary arithmetic could produce a different answer.

Another example is

-100

The correct output is

-100

Negative values are valid inputs, so the sign must be preserved exactly.

Approaches

A brute-force approach would be to decode the mathematical expression from the picture, evaluate it, and print the result. Since there is only one number involved, this already requires only constant work and is perfectly acceptable.

The key observation is that the expression in the image is mathematically equivalent to the variable itself. Once this is recognized, no computation remains. The answer is exactly the number that was read.

The brute-force method works because evaluating the expression gives the correct value. The simplification step removes even that tiny amount of computation by observing that the formula always collapses to a.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(1)	O(1)	Accepted
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

1. Read the integer a from the input.
2. Output a unchanged.

The reason this is correct is that the expression shown in the statement simplifies exactly to the original variable.

Why it works

The invariant is straightforward: the value represented by the formula is always equal to a. Since the algorithm outputs the same value that was read, it matches the mathematical quantity requested by the problem.

Python Solution

import sys
input = sys.stdin.readline

# solution
a = int(input())
print(a)

The program first reads the integer using fast input. Since there is only one value, no loops or auxiliary data structures are needed.

The only subtle point is preserving the sign. Converting with int() handles both positive and negative numbers automatically. Printing the value directly avoids any possibility of changing it accidentally.

Worked Examples

Example 1

Input:

1

Step	a	Output
Read input	1	-
Print answer	1	1

This trace shows the basic case. The number read from the input is returned unchanged.

Example 2

Input:

-7

Step	a	Output
Read input	-7	-
Print answer	-7	-7

This example confirms that negative numbers are handled correctly and that the sign is preserved.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only one integer is read and printed
Space	O(1)	No extra storage is required

The amount of work does not depend on the input value. The solution easily satisfies the limits.

Test Cases

# helper: run solution on input string, return output string
import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    a = int(input())
    return str(a)

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

# custom cases
assert run("0\n") == "0", "zero value"
assert run("-100\n") == "-100", "minimum input"
assert run("100\n") == "100", "maximum input"
assert run("-1\n") == "-1", "negative value"

Test input	Expected output	What it validates
0	0	Zero is preserved
-100	-100	Lower boundary
100	100	Upper boundary
-1	-1	Negative numbers are handled correctly

Edge Cases

A common source of mistakes is assuming that the picture requires additional computation. Consider

1

The algorithm reads a = 1 and immediately prints 1. Since the expression equals the variable itself, this is the correct answer.

Another important case is the smallest allowed value:

-100

The algorithm reads a = -100 and prints -100. No arithmetic is performed, so the sign and magnitude remain unchanged.

The same reasoning applies to all values between -100 and 100, inclusive. The answer is always exactly the input number.