CF 103409B - A Plus B Problem
This is the classic integer addition task framed in a competitive programming setting. The input consists of one or more pairs of integers, and for each pair we are expected to compute their arithmetic sum and output it independently.
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Solve time: 45s
Verified: yes
Solution
Problem Understanding
This is the classic integer addition task framed in a competitive programming setting. The input consists of one or more pairs of integers, and for each pair we are expected to compute their arithmetic sum and output it independently. There is no interaction between test cases, so each pair can be processed in isolation without storing any global state beyond the current line being read.
From a computational perspective, each test case requires only constant time work: reading two integers, performing one addition, and printing the result. Even if the number of test cases is large, the total complexity remains linear in the input size, which is optimal since every integer must be read at least once.
The main edge cases here are less about algorithmic structure and more about input format and integer range. A naive implementation might fail if it assumes only a single pair instead of multiple test cases. For example, if the input is:
3
1 2
-5 10
1000000000 1000000000
the correct output is:
3
5
2000000000
A common mistake is to read only the first line and ignore the remaining test cases, producing only 3. Another subtle issue is using a language type that cannot hold large sums safely, but in Python this is not a concern due to arbitrary precision integers.
Approaches
The brute-force interpretation of the problem is almost identical to the optimal solution. One could imagine treating each pair as a small computation problem: parse the numbers, compute their sum using standard arithmetic, and output it immediately. There is no meaningful way to simplify beyond this because addition itself is already O(1).
If we were to overthink the problem, we might try to store all pairs first and process them later, but that only increases memory usage without changing time complexity. Another unnecessary variant would be converting numbers into strings and simulating digit-by-digit addition, which would still be linear in the number of digits and strictly slower than native integer arithmetic.
The key observation is that the structure of the problem does not require any preprocessing, sorting, or dynamic programming. Each test case is independent, so streaming input directly into output is sufficient and optimal.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force (direct per test case addition) | O(T) | O(1) | Accepted |
| Optimal (streamed processing) | O(T) | O(1) | Accepted |
Algorithm Walkthrough
- Read the number of test cases, which determines how many independent addition operations we will perform. This allows us to structure the loop so that each pair is handled exactly once.
- For each test case, read two integers from input. This step is necessary because the problem guarantees that each computation is independent and self-contained.
- Compute the sum of the two integers using native arithmetic. This is the core operation, and it is constant time regardless of input magnitude in Python.
- Immediately output the computed sum before moving to the next test case. Streaming output avoids unnecessary storage and keeps memory usage constant.
Why it works
The correctness relies on the fact that addition is associative and independent across test cases. Each pair of integers forms a closed computation unit: no later operation depends on earlier results. Because we process each pair exactly once and apply the exact arithmetic operation required by the problem, the output sequence is guaranteed to match the input sequence of test cases.
Python Solution
import sys
input = sys.stdin.readline
def solve():
t = int(input())
out = []
for _ in range(t):
a, b = map(int, input().split())
out.append(str(a + b))
sys.stdout.write("\n".join(out))
if __name__ == "__main__":
solve()
The solution reads all test cases sequentially and accumulates results in a list for efficient output. Using sys.stdin.readline ensures fast input parsing, and joining at the end avoids repeated I/O overhead.
A subtle implementation detail is buffering output instead of printing line by line. While individual prints would still be correct, they can become slow when T is large due to repeated system calls. Accumulating results and writing once is the standard competitive programming optimization.
Worked Examples
Example 1
Input:
3
1 2
-5 10
7 7
| Step | a | b | sum | output so far |
|---|---|---|---|---|
| 1 | 1 | 2 | 3 | 3 |
| 2 | -5 | 10 | 5 | 3 5 |
| 3 | 7 | 7 | 14 | 3 5 14 |
This trace shows that each pair is processed independently and appended in order. The key property verified here is stability of output order.
Example 2
Input:
2
1000000000 1000000000
-100 -200
| Step | a | b | sum | output so far |
|---|---|---|---|---|
| 1 | 1000000000 | 1000000000 | 2000000000 | 2000000000 |
| 2 | -100 | -200 | -300 | 2000000000 -300 |
This confirms that the solution correctly handles both large positive integers and negative values without overflow or formatting issues.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(T) | Each test case requires constant time addition and output formatting |
| Space | O(1) | Only a fixed number of variables are used beyond the output buffer |
The runtime scales linearly with the number of test cases, which is optimal since every input line must be read and processed at least once.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
from __main__ import solve
sys.stdout = io.StringIO()
solve()
return sys.stdout.getvalue().strip()
# provided-style sample
assert run("3\n1 2\n-5 10\n7 7\n") == "3\n5\n14"
# single test case
assert run("1\n0 0\n") == "0"
# negative numbers
assert run("2\n-1 -1\n-5 2\n") == "-2\n-3"
# large numbers
assert run("2\n1000000000 1\n999999999 999999999\n") == "1000000001\n1999999998"
| Test input | Expected output | What it validates |
|---|---|---|
| single zero case | 0 | minimal boundary |
| negative values | -2 -3 | sign handling |
| large integers | 1000000001, 1999999998 | overflow safety and correctness |
Edge Cases
One edge case is when the input contains only a single test case. The algorithm still works because the loop executes exactly once and produces a single output line.
Another edge case is negative integers, for example input -5 3. The addition step remains unchanged since Python handles signed integers natively, and the output correctly reflects arithmetic sum -2.
A final edge case is very large integers close to typical 32-bit or 64-bit limits. Even for values like 10^9 + 10^9, the algorithm remains correct because Python’s integer type automatically expands to accommodate the result without overflow or precision loss.