CF 104545J - Joyful Feast of the Gods

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Tags: -
Solve time: 43s
Verified: yes

Solution

Problem Understanding

The story describes a character who adds money to a cafeteria account exactly once and then repeatedly spends from it on meals. Each meal has a fixed cost of two units of currency. After some unknown number of meals, we observe two values: how much money was added in that single recharge, and the remaining balance at the moment we check.

The task is to reconstruct how many meals were paid for after that recharge. The key hidden structure is that the only way the balance changes after the recharge is by repeated subtraction of a constant cost per meal, so the entire problem reduces to finding how much money disappeared from the account and dividing it by the price of one meal.

The input gives two integers, the recharge amount and the current balance. The output is the number of meals consumed after the recharge.

The constraints are extremely small, with both values bounded by 200 and guaranteed to have the same parity. This immediately rules out any need for simulation or search. Any solution that runs in constant time or linear time over the value range is already more than sufficient.

A naive but common mistake is to try reconstructing an unknown initial balance before the recharge. For example, one might assume the account started at zero, or try to guess pre-recharge history. That is unnecessary because the problem explicitly states there was only one recharge, and everything relevant happens after it.

Another potential confusion comes from the parity constraint. Without noticing that each meal costs exactly two, one might think parity is a red herring. In reality it guarantees that the difference between recharge and final balance is divisible by two, so the answer is always an integer.

Approaches

A brute-force interpretation would simulate meals one by one. Starting from the recharge amount, we repeatedly subtract two until we reach the final balance. This works because each operation directly corresponds to one meal, and we stop exactly when the observed balance is reached. However, even though the bounds are small here, this approach is conceptually wasteful since it performs one iteration per meal.

The key observation is that all meals have identical cost. Instead of simulating repeated subtraction, we can compute the total spent amount directly as the difference between what was added and what remains. That total spent amount must equal the number of meals multiplied by two. So the answer is obtained by a single division.

This reduces the problem from a step-by-step process into a direct arithmetic extraction.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Simulation	O(x)	O(1)	Acceptable but unnecessary
Direct Formula	O(1)	O(1)	Accepted

Algorithm Walkthrough

We compute how much money was used after the recharge and convert it into meal count.

Read the two integers, the recharge value and the current balance. These represent the starting amount after top-up and the observed remaining amount.
Compute the total spent amount by subtracting the final balance from the recharge amount. This difference corresponds exactly to all meals combined because there are no other operations affecting the balance.
Divide the spent amount by two to obtain the number of meals, since each meal costs exactly two units.
Output the result.

Why it works

After the single recharge, the only possible operation affecting the balance is paying for meals, each reducing the balance by exactly two. Therefore the total reduction in balance is exactly twice the number of meals. Since no other transactions occur, the difference between initial post-recharge balance and final balance uniquely determines the meal count. The parity condition guarantees this difference is always even, so integer division is valid without remainder.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    x, y = map(int, input().split())
    spent = x - y
    print(spent // 2)

if __name__ == "__main__":
    solve()

The implementation directly follows the derived formula. The subtraction x - y captures total expenditure after the recharge. There is no need for loops or simulation because every unit of spent money corresponds to a fixed number of meals.

The only subtlety is ensuring integer division is used. Since the problem guarantees identical parity, x - y is always even, so floor division safely produces the correct integer result.

Worked Examples

We trace the computation on two inputs.

Example 1

Input:

50 34

recharge x	balance y	spent (x - y)	meals (spent / 2)
50	34	16	8

The difference between recharge and remaining balance is 16. Since each meal costs 2, this corresponds to 8 meals. The trace confirms that the solution depends only on aggregate spending, not on any intermediate steps.

Example 2

Input:

131 47

recharge x	balance y	spent (x - y)	meals (spent / 2)
131	47	84	42

Here the total spent amount is 84, which decomposes into 42 meals of cost 2 each. This case demonstrates that the same logic holds for larger values without any change in structure.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only a constant number of arithmetic operations are performed
Space	O(1)	No auxiliary data structures are used

The constraints allow values up to 200, but the algorithm does not depend on their size. It performs a fixed sequence of operations regardless of input magnitude, making it trivially within limits.

Test Cases

# helper: run solution on input string, return output string
import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    x, y = map(int, sys.stdin.readline().split())
    spent = x - y
    return str(spent // 2)

# provided samples
assert run("50 34\n") == "8", "sample 1"
assert run("131 47\n") == "42", "sample 2"

# custom cases
assert run("0 0\n") == "0", "no recharge spent"
assert run("2 0\n") == "1", "single meal"
assert run("200 200\n") == "0", "full balance untouched"
assert run("200 0\n") == "100", "maximum spending case"

Test input	Expected output	What it validates
0 0	0	no spending after recharge
2 0	1	smallest non-trivial meal count
200 200	0	edge case where nothing is spent
200 0	100	maximum consumption boundary

Edge Cases

When the recharge equals the current balance, the difference is zero, so the algorithm outputs zero meals. This corresponds to a situation where no spending occurred after the recharge, and the subtraction step immediately produces zero.

When the balance drops to zero, all money from the recharge has been consumed. The algorithm computes x - 0, which equals the full recharge amount, and dividing by two yields the total number of meals. Since the parity constraint ensures x is even in this case, the division is always exact and produces a valid integer count.