CF 1843C - Sum in Binary Tree

CF 1843C - Sum in Binary Tree

Rating: 800
Tags: bitmasks, combinatorics, math, trees
Solve time: 1m 27s
Verified: yes

Solution

Problem Understanding

The problem describes an infinite complete binary tree where nodes are numbered sequentially layer by layer. The root has number 1, its left child is 2, right child is 3, and the pattern continues such that each node gets two children with the next available numbers. Given a vertex number n, we are asked to compute the sum of all node numbers along the path from the root to n.

The input consists of multiple test cases, each specifying a vertex number n which can be as large as $10^{16}$. This immediately rules out any solution that attempts to explicitly construct the tree or traverse it node by node, because a brute-force traversal could involve up to $10^{16}$ operations, far exceeding feasible limits for a 1-second runtime.

Non-obvious edge cases include the smallest possible node, n = 1. In this case the path contains only the root, and the sum should be 1. Another subtlety arises when n is a power of two or just after a power of two. The numbering of nodes by layers aligns naturally with binary representations, and this is key for an efficient solution. A careless approach that tries to simulate children assignment sequentially would fail for large n.

Approaches

The brute-force approach would generate the tree up to node n, or maintain a parent pointer for each node. We could then follow parent pointers back from n to 1 to sum the path. This works for small n but becomes infeasible for n close to $10^{16}$, as it would require an enormous number of insertions or memory allocations.

The key insight comes from observing the structure of the tree. Every node in this infinite binary tree can be associated with a binary representation that encodes the path from the root. Specifically, if we number the root as 1, its left and right children as 2 and 3, and continue layer by layer, then each step downward corresponds to choosing the left or right child, which is equivalent to a bit in n's binary form. More concretely, if we trace the path from n to the root, each move to a parent can be computed with integer division by 2, and summing these values gives the desired result. This reduces the problem to repeatedly halving n and accumulating the sum until we reach 1.

This approach works because the tree numbering corresponds exactly to the sequence of numbers produced by a breadth-first traversal of a complete binary tree. Moving from a node to its parent is n // 2, and this can be repeated until we reach the root. This guarantees a logarithmic number of steps proportional to the number of bits in n.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n)	O(n)	Too slow for large n
Optimal	O(log n)	O(1)	Accepted

Algorithm Walkthrough

1. Read the number of test cases, t. For each test case, read the vertex number n.
2. Initialize a variable ans to 0. This will hold the sum of nodes along the path.
3. While n is greater than 0, add n to ans and update n to n // 2. Each division by 2 moves one step up the tree.
4. Once n reaches 0, the path is fully traversed. Output ans.

Why it works: The invariant is that at every iteration, n represents the current node on the path to the root. Adding n to ans accumulates the sum, and dividing by 2 moves us to its parent. This guarantees we include every node in the path exactly once, terminating at the root.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n = int(input())
    ans = 0
    while n > 0:
        ans += n
        n //= 2
    print(ans)

The solution reads input efficiently using sys.stdin.readline, which is necessary for up to $10^4$ test cases. The while loop ensures that we trace the path from n to the root in logarithmic time. Using integer division guarantees correct parent computation, and Python's arbitrary-precision integers handle very large values of n.

Worked Examples

Example 1: n = 3

Step	n	ans
1	3	0
2	1	3
3	0	4

The path from 1 → 3 is correctly summed as 1 + 3 = 4.

Example 2: n = 10

Step	n	ans
1	10	0
2	5	10
3	2	15
4	1	17
5	0	18

The path from 1 → 2 → 5 → 10 is summed correctly as 1 + 2 + 5 + 10 = 18.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(log n) per test case	Each iteration divides n by 2, and the number of iterations is equal to the number of bits in n.
Space	O(1)	Only a single integer accumulator is required; no tree or array structures are needed.

With t up to $10^4$ and n up to $10^{16}$, the total number of operations is about $10^4 * 54$ since 54 bits suffice for $10^{16}$, which is well within time limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    # solution code
    t = int(input())
    for _ in range(t):
        n = int(input())
        ans = 0
        while n > 0:
            ans += n
            n //= 2
        print(ans)
    return output.getvalue().strip()

# provided samples
assert run("6\n3\n10\n37\n1\n10000000000000000\n15\n") == "4\n18\n71\n1\n19999999999999980\n26"

# custom cases
assert run("1\n1\n") == "1", "root node"
assert run("1\n2\n") == "3", "small left child"
assert run("1\n3\n") == "4", "small right child"
assert run("1\n16\n") == "31", "power of 2"
assert run("1\n17\n") == "48", "just above power of 2"

Test input	Expected output	What it validates
1	1	Path sum for root node
2	3	Smallest left child
3	4	Smallest right child
16	31	Power of two node, full path depth
17	48	Node just beyond power of two, checks path calculation correctness

Edge Cases

For n = 1, the algorithm sets ans = 0, adds 1, then divides by 2 to get 0. The output is 1, as expected. For very large n, such as 10^{16}, Python handles the integer size seamlessly, and the loop executes at most 54 iterations due to the binary length of n, confirming correctness for the maximum constraint. The division by 2 correctly traces parents for both even and odd n, ensuring no off-by-one errors.