CF 104493E - Sad Teacher
We are given a single student’s marks in two subjects: Physics and Chemistry. Each test case provides two integers representing these scores, and the task is simply to compute the student’s total score by adding them together.
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Solve time: 1m 4s
Verified: yes
Solution
Problem Understanding
We are given a single student’s marks in two subjects: Physics and Chemistry. Each test case provides two integers representing these scores, and the task is simply to compute the student’s total score by adding them together.
The input is not describing a structure like an array or graph, but rather a direct aggregation problem where each pair of numbers is independent. The output for each pair is the arithmetic sum of those two values.
The constraints allow each number to be as large as $10^{18}$. This immediately rules out any approach concerns related to algorithmic complexity, since there is no iteration over large datasets or combinatorial structure. The only real consideration is whether the chosen numeric type can safely hold the sum without overflow. In languages with fixed-width integers, this would require 128-bit integers or careful handling. In Python, integers are unbounded, so this is naturally safe.
There are no subtle structural edge cases in terms of ordering or multiple interpretations of the input. The only situations that can cause incorrect behavior are implementation mistakes such as reading input incorrectly, forgetting to convert strings to integers, or using a type that overflows.
A typical pitfall appears when someone assumes multiple test cases or extra formatting. For example, interpreting the input incorrectly like:
Input:
70 80
If mistakenly parsed as a single string without splitting, one might attempt concatenation and produce "7080" instead of 150, which is incorrect. The correct interpretation is numeric addition.
Approaches
The most direct way to solve this problem is to read both integers and compute their sum. This brute-force interpretation already matches the optimal approach because there is no hidden structure to exploit and no repeated computation.
If one were to think in overly general terms, a “brute-force” approach might involve converting numbers into strings, simulating digit-by-digit addition, and managing carry manually. That would correctly compute the sum but is unnecessary overhead for this problem size. It would also introduce complexity where none is needed.
The key observation is that the problem is purely arithmetic addition of two integers within a safe range for modern integer types. Since Python supports arbitrary precision integers, the optimal solution is identical in structure to the naive conceptual solution but implemented directly using the + operator.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Digit-by-digit manual addition | O(d) | O(d) | Correct but unnecessary |
| Direct integer addition | O(1) | O(1) | Accepted |
Algorithm Walkthrough
- Read the entire input line from standard input. The problem guarantees two integers, so a single read is sufficient.
- Split the input into two tokens representing the two marks.
- Convert both tokens into integers so arithmetic addition is performed rather than string concatenation.
- Compute the sum of the two integers using the built-in addition operator.
- Output the resulting sum as a single integer.
Why it works
The correctness follows from the fact that the problem definition is exactly integer addition. There are no transformations, constraints, or hidden conditions. As long as both inputs are correctly parsed as integers, Python’s addition operator produces the mathematically correct result, and the output directly matches the required total score.
Python Solution
import sys
input = sys.stdin.readline
def solve():
data = input().strip().split()
a = int(data[0])
b = int(data[1])
print(a + b)
if __name__ == "__main__":
solve()
The solution reads a single line, splits it into two components, and converts them into integers. The use of strip().split() ensures robustness against extra whitespace or trailing newline characters. The addition a + b is performed using Python’s arbitrary precision integers, so even values up to $10^{18}$ are safely handled without overflow.
A common mistake in similar problems is forgetting the integer conversion step and accidentally concatenating strings. Another subtle issue in other languages would be integer overflow, but Python avoids that entirely.
Worked Examples
Example 1
Input:
70 80
| Step | Tokens | a | b | a + b |
|---|---|---|---|---|
| Parse input | ["70", "80"] | - | - | - |
| Convert | ["70", "80"] | 70 | 80 | - |
| Compute | - | 70 | 80 | 150 |
The computation is straightforward integer addition. The final result is 150, which matches the expected output.
Example 2
Input:
4 5
| Step | Tokens | a | b | a + b |
|---|---|---|---|---|
| Parse input | ["4", "5"] | - | - | - |
| Convert | ["4", "5"] | 4 | 5 | - |
| Compute | - | 4 | 5 | 9 |
This confirms that even for small inputs, the logic remains identical and no special casing is required.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(1) | Only constant-time parsing and one addition operation |
| Space | O(1) | Only two integer variables are stored |
The constraints allow values up to $10^{18}$, but since there is no iteration or data structure scaling with input size, the runtime remains constant. Memory usage is also constant because only two integers are stored at any time.
This easily fits within both time and memory limits for any standard competitive programming environment.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
from sys import stdout
old_stdout = sys.stdout
sys.stdout = io.StringIO()
solve()
output = sys.stdout.getvalue()
sys.stdout = old_stdout
return output.strip()
def solve():
data = sys.stdin.readline().strip().split()
a = int(data[0])
b = int(data[1])
print(a + b)
# provided samples
assert run("70 80\n") == "150", "sample 1"
assert run("4 5\n") == "9", "sample 2"
# custom cases
assert run("1 1\n") == "2", "minimum values"
assert run("1000000000000000000 1000000000000000000\n") == "2000000000000000000", "maximum values"
assert run("0 123\n") == "123", "zero boundary case"
assert run("999999999999999999 1\n") == "1000000000000000000", "carry boundary case"
| Test input | Expected output | What it validates |
|---|---|---|
| 1 1 | 2 | Minimum valid input handling |
| max values | 2000000000000000000 | Large integer correctness |
| 0 123 | 123 | Zero boundary behavior |
| 999...9 1 | 1000000000000000000 | Carry across digit boundary |
Edge Cases
One edge case is the maximum constraint input where both numbers are at $10^{18}$. In such a case, the input is:
1000000000000000000 1000000000000000000
The algorithm reads both values as integers, stores them without truncation, and computes their sum as $2 \times 10^{18}$. Python handles this cleanly because integer precision is unbounded, so no overflow or wrapping occurs.
Another edge case is when one of the values is zero:
0 123
The parsing step converts "0" into integer 0, and the sum remains unchanged as 123. This verifies that no special-case logic is incorrectly applied to zero values.
A final edge scenario is where the addition results in a carry across many digits, such as:
999999999999999999 1
The algorithm converts both strings into integers and performs arithmetic addition. Internally, Python manages the carry propagation automatically, producing the correct result 1000000000000000000 without any manual intervention.