CF 1351A - A+B (Trial Problem)

Rating: 800
Tags: implementation
Solve time: 2m 17s
Verified: yes

Solution

Problem Understanding

The task asks us to add pairs of integers. Each test case consists of two integers, and for each pair, we need to compute their sum. The input first tells us how many pairs there are, then each subsequent line contains a pair. The output is a sequence of sums, one per line, corresponding to each pair.

The constraints are straightforward but worth noting. Each integer is between -1000 and 1000, which means integer overflow is not a concern in Python. The number of test cases can be as large as 10,000, which implies that any solution iterating through all test cases once is efficient enough, as Python can comfortably handle 10,000 simple operations in under a second. Negative numbers introduce the possibility of sums that are zero or negative, so the solution must not assume inputs are strictly positive.

Non-obvious edge cases include pairs where one or both numbers are zero, where the sum crosses zero from positive to negative or vice versa, and where both numbers are at the bounds (-1000 or 1000). For example, the pair -1000 1000 should output 0. A careless implementation that neglects negatives might produce the wrong sign.

Approaches

The brute-force approach is to read each line, parse the two integers, compute their sum, and print the result immediately. This works because summing two integers is an O(1) operation, and iterating over 10,000 test cases is efficient in Python. There is no hidden complexity in reading and processing the inputs sequentially.

There is no more optimal asymptotic approach because we must process every test case to compute its sum. The only improvement might be to batch input reading for speed using sys.stdin.readline, which avoids Python’s slower default input. This small optimization ensures that even at the maximum input size, the program runs comfortably within time limits.

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

Algorithm Walkthrough

Read the number of test cases t. This tells us how many pairs we need to process.
Iterate t times. In each iteration:
Read a line containing two integers, a and b.
Convert the line into integers. The conversion is necessary because input is a string.
Compute the sum a + b.
Print the sum immediately.
Repeat for all test cases.

The key invariant is that for each iteration, a and b are valid integers within the constraints. Summing them produces the correct answer, and printing immediately maintains the output order.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    a, b = map(int, input().split())
    print(a + b)

The first line reads the number of test cases. The loop runs exactly t times, reading each pair, converting to integers, summing, and printing. Using sys.stdin.readline avoids slow input operations on large test files. map(int, input().split()) ensures both values are parsed correctly even if negative. Printing inside the loop preserves order without extra storage.

Worked Examples

Sample Input 1

Iteration	a	b	a+b
1	1	5	6
2	314	15	329
3	-99	99	0
4	123	987	1110

This trace confirms the algorithm correctly handles positive, negative, and zero sums, printing them in order.

Custom Input

Iteration	a	b	a+b
1	0	0	0
2	-1000	1000	0
3	500	-200	300

This demonstrates the algorithm correctly processes boundary values and mixed signs.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t)	Each test case requires O(1) operations: reading two numbers, summing, printing.
Space	O(1)	No additional data structures are required; only current `a` and `b` are stored.

Given t ≤ 10,000, this solution runs in a fraction of a second and uses negligible memory.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    t = int(input())
    for _ in range(t):
        a, b = map(int, input().split())
        print(a + b)
    return output.getvalue().strip()

# provided sample
assert run("4\n1 5\n314 15\n-99 99\n123 987") == "6\n329\n0\n1110", "sample 1"

# minimum inputs
assert run("1\n0 0") == "0", "minimum input"

# maximum inputs
assert run("2\n1000 1000\n-1000 -1000") == "2000\n-2000", "maximum absolute values"

# mixed signs
assert run("3\n-500 500\n-200 100\n0 -1") == "0\n-100\n-1", "mixed signs"

# single test case
assert run("1\n123 456") == "579", "single case"

Test input	Expected output	What it validates
1 0 0	0	minimum values
2 1000 1000 \n -1000 -1000	2000\n-2000	maximum absolute values
3 -500 500 \n -200 100 \n 0 -1	0\n-100\n-1	handling mixed signs and zeros
1 123 456	579	correctness for a single test case

Edge Cases

For the pair -1000 1000, the algorithm reads a=-1000, b=1000, computes a+b=0, and prints 0. The invariant that each read number is within bounds guarantees no overflow. The sum is correctly handled regardless of sign, demonstrating the algorithm’s robustness at input extremes. Similarly, for 0 0, the sum is 0, confirming correct behavior for minimal inputs.