CF 102697048 - Basic Math

CF 102697048 - Basic Math

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Solve time: 51s
Verified: yes

Solution

Problem Understanding

The task is to evaluate a single arithmetic equation containing integers connected by addition and subtraction. The equation ends with an equals sign, and the required output keeps the original equation intact while appending its numerical result.

For example, the input 1 + 5 + 7 - 2 = represents the calculation (1+5+7-2), whose value is (11). The required output is 1 + 5 + 7 - 2 = 11. The problem statement does not give a separate numeric bound on the size of the equation, but the time limit is one second and the expression is an ordinary input string. That makes a linear scan the natural target. If the expression has (L) characters, an (O(L)) solution performs a constant amount of work per character, while repeatedly rescanning the expression can grow to (O(L^2)).

The main edge cases come from the fact that subtraction is not commutative and that the equals sign is part of the input format. For 8 - 3 - 2 =, the correct output is 8 - 3 - 2 = 3. A careless implementation that treats all numbers as positive would produce (13), and an implementation that interprets subtraction as a single operation could produce the wrong grouping.

For 5 - 5 + 5 - 5 =, the correct output is 5 - 5 + 5 - 5 = 0. This catches implementations that accidentally add every parsed number and only inspect the operators afterward.

For the smallest useful equation, 7 =, the correct output is 7 = 7. There is no operator to process, so the first number must directly become the result.

Approaches

A straightforward approach is to repeatedly search through the remaining expression for the next number and its following operator. This is correct because the expression contains only addition and subtraction, so evaluating terms from left to right gives exactly the arithmetic result. The problem with this implementation is unnecessary rescanning. If every search starts again from the beginning of the unprocessed portion, an expression containing (L) characters can cause roughly (L+(L-1)+\cdots+1=O(L^2)) character inspections. For a sufficiently long input, that quadratic behavior is avoidable work.

The key observation is that addition and subtraction have no precedence differences. Once a number and its sign have been read, its contribution to the final answer is completely determined. We never need to revisit an earlier character. A single left-to-right scan can accumulate the current number, and whenever a plus or minus sign is encountered, the completed number can immediately be added or subtracted from the answer.

The equals sign does not participate in the calculation. It simply marks the end of the arithmetic expression. We can stop processing when we reach it and keep the original input line for the required output.

The difference between the two approaches is thus not a more complicated mathematical technique. It is recognizing that the expression is a stream of independent signed terms, so each character needs to be inspected only once.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(L²)	O(L)	Too slow for large expressions
Optimal	O(L)	O(1)	Accepted

Here (L) denotes the length of the input equation.

Algorithm Walkthrough

1. Read the complete equation as a string and preserve it exactly for the final output. The original formatting is part of what the problem asks us to print.
2. Scan the equation from left to right while maintaining the accumulated answer, the current number, and the sign that will be applied to that number. Starting with a positive sign handles the first number naturally.
3. When a digit is encountered, extend the current number with that digit. For example, reading 42 changes the current number from 4 to 42.
4. When + or - is encountered, apply the previously stored sign to the completed current number, then reset the current number to zero and remember the newly encountered sign. Processing the previous number before changing the sign avoids associating it with the wrong operator.
5. When = is encountered, apply the final number using its stored sign and stop. The equation is complete at this point.
6. Print the untouched original equation, followed by a space and the calculated answer. This separates evaluating the expression from formatting the required output.

Why it works

At every point during the scan, the accumulated answer equals the value of every complete term that appears before the current number. The current number contains exactly the digits of the next term, and the stored sign is exactly the operator immediately preceding that term. When an operator or the equals sign is reached, applying that sign adds precisely the next term's contribution to the accumulated answer. Since every term is processed exactly once and the final term is processed when = is reached, the accumulated value at the end is exactly the value of the entire equation.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    equation = input().rstrip("\n")

    answer = 0
    number = 0
    sign = 1

    for ch in equation:
        if ch.isdigit():
            number = number * 10 + ord(ch) - ord('0')
        elif ch == '+' or ch == '-':
            answer += sign * number
            number = 0
            sign = 1 if ch == '+' else -1
        elif ch == '=':
            break

    answer += sign * number

    print(equation + " " + str(answer))

if __name__ == "__main__":
    solve()

The equation variable stores the original line so that the output can reproduce it without rebuilding the expression. Using rstrip("\n") removes only the line ending, rather than removing arbitrary whitespace from the equation.

The scanner treats every digit as part of the current integer. The expression may contain multi-digit values, so replacing number with the current digit would be incorrect. The update number = number * 10 + digit constructs the decimal integer in the usual way.

When an operator is found, the code first adds the current number using the old sign. Only afterward does it change the sign. This order is essential for expressions such as 10 - 3 + 4 =.

The equals sign terminates evaluation because everything after it would be outside the arithmetic expression. The final answer += sign * number is necessary because the last number has no operator after it to trigger processing.

Python integers do not overflow for ordinary integer input sizes, so no special overflow handling is required.

Worked Examples

Sample 1

For 1 + 5 + 7 - 2 =, the scanner processes each term as follows.

Character reached	Current number	Sign	Answer
1	1	+1	0
+	0	+1	1
5	5	+1	1
+	0	+1	6
7	7	+1	6
-	0	-1	13
2	2	-1	13
=	2	-1	13
final term	0	-1	11

The result is 11, so the output is 1 + 5 + 7 - 2 = 11. The trace demonstrates why the sign must be changed only after the preceding number has been added to the answer.

Custom example

Consider 20 - 7 - 3 + 10 =.

Character reached	Current number	Sign	Answer
20	20	+1	0
-	0	-1	20
7	7	-1	20
-	0	-1	13
3	3	-1	13
+	0	+1	10
10	10	+1	10
=	10	+1	10
final term	0	+1	20

The final value is 20, giving 20 - 7 - 3 + 10 = 20. This example confirms that consecutive subtraction operations are handled independently, rather than being combined into an incorrect expression.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(L)	Every character of the equation is inspected once.
Space	O(1) auxiliary space	Only the answer, current number, sign, and a few temporary values are maintained.

The solution scales linearly with the length of the input. Even if the equation contains a very large number of terms, the program performs only a constant amount of work for each character, which is the appropriate complexity for a one-second limit.

Test Cases

import sys
import io

def solve():
    equation = input().rstrip("\n")

    answer = 0
    number = 0
    sign = 1

    for ch in equation:
        if ch.isdigit():
            number = number * 10 + ord(ch) - ord('0')
        elif ch == '+' or ch == '-':
            answer += sign * number
            number = 0
            sign = 1 if ch == '+' else -1
        elif ch == '=':
            break

    answer += sign * number
    print(equation + " " + str(answer))

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

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

    try:
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# Provided sample
assert run("1 + 5 + 7 - 2 =\n") == "1 + 5 + 7 - 2 = 11\n", "sample 1"

# Minimum-size equation
assert run("7 =\n") == "7 = 7\n", "single number"

# All values cancel
assert run("5 - 5 + 5 - 5 =\n") == "5 - 5 + 5 - 5 = 0\n", "all-equal values"

# Multi-digit values and repeated subtraction
assert run("20 - 7 - 3 + 10 =\n") == "20 - 7 - 3 + 10 = 20\n", "multi-digit terms"

# Boundary-style large expression
terms = ["1"] * 100000
large_equation = " + ".join(terms) + " =\n"
assert run(large_equation) == large_equation.rstrip("\n") + " 100000\n", "large input"

Test input	Expected output	What it validates
7 =	7 = 7	Minimum-size equation and final-term handling
5 - 5 + 5 - 5 =	5 - 5 + 5 - 5 = 0	Repeated addition and subtraction with equal values
20 - 7 - 3 + 10 =	20 - 7 - 3 + 10 = 20	Multi-digit numbers and consecutive subtraction
100000 copies of 1 joined by +	Same equation followed by 100000	Large input and linear scanning

Edge Cases

For the single-number case 7 =, the scanner reads 7 into number, reaches =, and then adds the final term using the initial positive sign. The result is 7, so the output is 7 = 7. No operator is required for the algorithm to work.

For 8 - 3 - 2 =, the first - causes 8 to be added and changes the sign to negative. The second - adds -3 and keeps the sign negative for the next term. The final 2 is consequently subtracted, producing 3. The output is 8 - 3 - 2 = 3.

For 5 - 5 + 5 - 5 =, the answer evolves as (5), (0), (5), and finally (0). Every operator applies only to the number immediately following it, while the accumulator contains all terms already completed.

For a large expression containing 100000 terms such as 1 + 1 + 1 + ... + 1 =, the scanner still visits each character once. The answer becomes 100000, and there is no nested loop or repeated substring search that could turn the work into quadratic time.