CF 1977A - Little Nikita

CF 1977A - Little Nikita

Rating: 800
Tags: math
Solve time: 1m 32s
Verified: yes

Solution

Problem Understanding

Nikita has a tower that starts empty, and he can perform exactly one of two operations per move: either place one cube on top or remove one cube from the top. The problem asks whether, after performing exactly n moves, the tower can have exactly m cubes. Each test case provides the number of moves n and the desired height m. The output is "Yes" if this is possible, and "No" otherwise.

The constraints are small: both n and m are at most 100, and there are at most 100 test cases. This means any algorithm with complexity roughly O(n) per test case will run comfortably within the time limit. The problem is essentially about parity and reachability rather than exhaustive simulation.

Non-obvious edge cases occur when m is greater than n or when m and n have different parity. For example, if n = 2 and m = 3, there is no way to reach three cubes in only two moves. Similarly, if n = 3 and m = 2, the moves must leave the tower at height 2 after three operations, but since each move changes the height by exactly 1, the parity of n and m must match for it to be possible.

Approaches

A brute-force approach would attempt to simulate all sequences of adding and removing cubes. For each move, you could recursively try adding or removing a cube and check whether the final height matches m. This works for very small inputs but has exponential complexity: each of n moves has two choices, so there are 2^n possible sequences. Even with n = 100, this is infeasible.

The key insight is that the only factors limiting the final tower height are the total number of moves n and the parity of m. Every move changes the tower height by exactly one, so after n moves, the height difference from the starting point (zero) must be a number of the same parity as n. Additionally, the maximum height achievable is n if Nikita only adds cubes. Therefore, the conditions for reaching exactly m cubes are that m is at most n and m has the same parity as n. This reduces the problem to a simple check without simulation.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(2^n)	O(n)	Too slow
Optimal	O(1) per test case	O(1)	Accepted

Algorithm Walkthrough

1. Read the number of test cases t.
2. For each test case, read the integers n and m.
3. Check if m is less than or equal to n. If not, output "No" and continue.
4. Check if the parity of n and m matches, meaning (n - m) % 2 == 0. If it does, output "Yes"; otherwise, output "No".

The reason this works is that each move changes the tower height by exactly 1, so the difference between n and m must be an even number to reach exactly m after n moves. The condition m <= n ensures that Nikita does not need more moves than allowed to reach the desired height.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n, m = map(int, input().split())
    if m <= n and (n - m) % 2 == 0:
        print("Yes")
    else:
        print("No")

The code reads input efficiently using sys.stdin.readline. For each test case, it immediately checks the two conditions: m <= n ensures we have enough moves to reach the height, and (n - m) % 2 == 0 ensures that the number of up and down moves can be arranged to reach exactly m. This avoids unnecessary loops or recursion.

Worked Examples

Example 1: n = 3, m = 3

n	m	n - m	parity check	Result
3	3	0	0 % 2 == 0	Yes

The difference is zero, even, and m <= n, so reaching 3 cubes is possible by adding a cube on each move.

Example 2: n = 2, m = 4

n	m	n - m	parity check	Result
2	4	-2	-2 % 2 == 0	No

Although the parity check is satisfied, m > n makes it impossible to reach height 4 in only 2 moves.

These traces show that both conditions are necessary and sufficient.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t)	Each test case is evaluated in constant time.
Space	O(1)	No additional memory beyond input storage.

With t up to 100 and n, m up to 100, the solution runs comfortably in under 1 second and uses negligible memory.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    sys.stdout = out
    t = int(input())
    for _ in range(t):
        n, m = map(int, input().split())
        if m <= n and (n - m) % 2 == 0:
            print("Yes")
        else:
            print("No")
    return out.getvalue().strip()

# provided samples
assert run("3\n3 3\n2 4\n5 3\n") == "Yes\nNo\nYes", "sample 1"

# custom cases
assert run("4\n1 1\n2 2\n2 1\n100 100\n") == "Yes\nYes\nNo\nYes", "edge and max"
assert run("2\n1 2\n3 0\n") == "No\nYes", "parity and zero"
assert run("3\n10 10\n10 0\n10 5\n") == "Yes\nYes\nNo", "all-even, all-zero, mismatched parity"

Test input	Expected output	What it validates
1 1, 2 2, 2 1, 100 100	Yes, Yes, No, Yes	Minimum size, exact matches, impossible height, maximum size
1 2, 3 0	No, Yes	Parity mismatch, reaching zero cubes
10 10, 10 0, 10 5	Yes, Yes, No	Even maximum, zero, and impossible parity

Edge Cases

When m = 0 and n is even, the algorithm correctly outputs "Yes" because Nikita can add and remove cubes alternately to end with an empty tower. For example, n = 4, m = 0 results in (n - m) % 2 = 0 and m <= n, so the output is "Yes".

When m > n, the algorithm immediately outputs "No", avoiding incorrect assumptions about negative moves. For example, n = 3, m = 5 produces "No" even though the parity matches, because the height is unreachable.

When n and m differ in parity, the algorithm outputs "No", catching the subtle case where a sequence of ±1 moves cannot reach the target due to parity mismatch, such as n = 5, m = 2.