CF 1532A - A+B (Trial Problem)
We are given a sequence of independent queries. Each query consists of two integers, and for every pair we must compute their arithmetic sum and output it immediately. There is no dependency between test cases.
CF 1532A - A+B (Trial Problem)
Rating: -
Tags: *special
Solve time: 2m 54s
Verified: yes
Solution
Problem Understanding
We are given a sequence of independent queries. Each query consists of two integers, and for every pair we must compute their arithmetic sum and output it immediately.
There is no dependency between test cases. Each pair can be processed in isolation, so the problem reduces to repeatedly applying a single constant-time operation.
The constraints allow up to 10,000 test cases, with each integer ranging between -1000 and 1000. This means the magnitude of values is small enough that integer overflow is not a concern in Python, and even in lower-level languages 32-bit integers are sufficient. The dominant factor is the number of operations across all test cases.
A naive interpretation might suggest building intermediate structures or storing all inputs before processing, but nothing in the problem requires that. Any approach that does more than constant work per test case would still pass here, but it would be unnecessary overhead.
Edge cases are minimal but still worth acknowledging. A common mistake is mishandling negative values or forgetting to output a newline per test case.
For example, input:
1
-1000 1000
Correct output is:
0
A careless implementation might attempt string concatenation instead of numeric addition and produce -10001000, which is incorrect because it treats the inputs as strings rather than integers.
Approaches
The brute-force interpretation is to read each test case, compute the sum using standard arithmetic, and print it. Since each operation is O(1), the total work is linear in the number of test cases.
There is no structure to exploit beyond direct computation. No prefix reuse, no sorting, no graph traversal. The problem is essentially a sanity check that the input pipeline and output formatting are correct.
The only meaningful “optimization” is ensuring fast input parsing and avoiding unnecessary conversions or string operations inside the loop. The optimal solution is therefore identical in algorithmic structure to the brute-force approach, differing only in implementation cleanliness and efficiency of I/O.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force (direct summation per test case) | O(t) | O(1) | Accepted |
| Optimal (fast I/O + direct summation) | O(t) | O(1) | Accepted |
Algorithm Walkthrough
-
Read the integer
t, which defines how many independent additions we must perform. This determines the number of iterations in the main loop. -
For each test case, read two integers
aandbfrom input. The parsing step ensures we interpret the values as numbers, not strings, so arithmetic behaves correctly. -
Compute the sum
a + b. This is the only transformation required per test case and directly corresponds to the problem statement. -
Print the result immediately, ensuring each output appears on its own line to preserve the required format.
Why it works
Each test case is independent and defines a single arithmetic expression. The addition operation is deterministic and has no side effects. Because we compute and output each result exactly once, the algorithm preserves a one-to-one mapping between inputs and outputs. There are no hidden states or interactions between test cases, so correctness follows directly from the correctness of integer addition.
Python Solution
import sys
input = sys.stdin.readline
def main():
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__":
main()
The solution reads input using fast buffered I/O, which is important when handling up to 10,000 lines. Each line is split into two integers, and their sum is appended to an output list. The decision to accumulate results before printing avoids repeated costly write operations.
A common mistake is printing inside the loop using print, which is still acceptable here but slower in tight constraints. Another potential issue is forgetting to convert input strings to integers, which would result in string concatenation instead of numeric addition.
Worked Examples
Example 1
Input:
4
1 5
314 15
-99 99
123 987
| Test case | a | b | a + b | Output |
|---|---|---|---|---|
| 1 | 1 | 5 | 6 | 6 |
| 2 | 314 | 15 | 329 | 329 |
| 3 | -99 | 99 | 0 | 0 |
| 4 | 123 | 987 | 1110 | 1110 |
This trace confirms that each pair is handled independently and that negative values cancel correctly when they are additive inverses.
Example 2
Input:
3
0 0
-1000 -1000
42 -17
| Test case | a | b | a + b | Output |
|---|---|---|---|---|
| 1 | 0 | 0 | 0 | 0 |
| 2 | -1000 | -1000 | -2000 | -2000 |
| 3 | 42 | -17 | 25 | 25 |
This demonstrates that the algorithm handles boundary values and mixed signs without any special casing.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(t) | Each test case performs a single addition and constant-time I/O parsing |
| Space | O(t) | Output storage holds one string per test case before printing |
The constraints allow up to 10,000 operations, and each operation is constant time, so the solution runs comfortably within limits. Memory usage is also minimal since only the output list is stored.
Test Cases
import sys, io
def solve():
input = sys.stdin.readline
t = int(input())
out = []
for _ in range(t):
a, b = map(int, input().split())
out.append(str(a + b))
return "\n".join(out)
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
return solve()
# provided samples
assert run("4\n1 5\n314 15\n-99 99\n123 987\n") == "6\n329\n0\n1110"
# minimum input
assert run("1\n0 0\n") == "0"
# all negative
assert run("2\n-1 -2\n-1000 -1000\n") == "-3\n-2000"
# mixed signs
assert run("3\n1 -1\n10 -3\n-7 20\n") == "0\n7\n13"
# boundary values
assert run("2\n1000 1000\n-1000 1000\n") == "2000\n0"
| Test input | Expected output | What it validates |
|---|---|---|
| single zero pair | 0 | minimum input handling |
| all negatives | -3, -2000 | sign correctness |
| mixed signs | 0, 7, 13 | arithmetic correctness across cases |
| boundary values | 2000, 0 | constraint limits |
Edge Cases
The simplest edge case is when both numbers are zero. Input:
1
0 0
The algorithm reads both integers, computes 0 + 0, and outputs 0. There is no special branch required, and the same logic applies as in all other cases.
Another case involves extreme values within the constraint range, such as:
1
-1000 1000
Here the sum is exactly zero. Since the algorithm performs direct integer addition, there is no risk of overflow or precision loss in Python, and the output remains correct without any additional handling.