CF 401A - Vanya and Cards
Vanya has found a subset of his playing cards, each card labeled with an integer between $-x$ and $x$. He wants the total sum of all his cards to be zero. We need to determine the minimum number of additional cards he should find to achieve that zero sum.
Rating: 800
Tags: implementation, math
Solve time: 4m 15s
Verified: yes
Solution
Problem Understanding
Vanya has found a subset of his playing cards, each card labeled with an integer between $-x$ and $x$. He wants the total sum of all his cards to be zero. We need to determine the minimum number of additional cards he should find to achieve that zero sum.
The input gives two numbers, $n$ and $x$, representing the number of cards he has found and the maximum absolute value allowed on a card. Then a list of $n$ integers describes the values of the found cards. The output is a single integer: the minimum number of new cards required to balance the sum to zero.
The constraints are small: $n \le 1000$ and $x \le 1000$. This allows algorithms that run in linear time relative to $n$. The largest sum in absolute value Vanya could have from the found cards is $1000 \times 1000 = 10^6$, which is safely handled with standard integer types.
A subtle case is when the sum of found cards is already divisible by $x$ or smaller than $x$. For instance, if Vanya has found one card with value 7 and $x = 3$, he cannot balance it with a single card, so he must use multiple cards. Careless solutions might assume the minimum number is always 1 if the sum is non-zero, which would be incorrect.
Approaches
The naive approach is to try all combinations of additional cards between $-x$ and $x$ until the total sum becomes zero. This is correct in principle because we can always choose enough cards to compensate for any sum, but it is impractical. For example, if the sum is 1000 and $x = 1$, we would need to explore sequences of up to 1000 cards, which is combinatorially explosive.
The key insight is to note that the only thing that matters is the absolute sum of the found cards. We have infinite cards of any value from $-x$ to $x$, so the problem reduces to: given a target sum $S$, how many cards with absolute value at most $x$ are needed to reach $S$? The minimum number of cards is simply the ceiling of $|S| / x$. This works because we can always pick the largest allowed card repeatedly in the direction opposite to the sum until we reach zero.
This transforms the problem from a combinatorial search into a simple arithmetic calculation, which runs in $O(n)$ to compute the sum of the input and $O(1)$ to calculate the number of required cards.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force (try all sequences) | Exponential | O(?) | Too slow |
| Optimal (sum & ceiling division) | O(n) | O(1) | Accepted |
Algorithm Walkthrough
- Read the integers $n$ and $x$. These represent the number of found cards and the maximum card value. We will need $x$ to determine the largest single-card adjustment we can make.
- Read the array of $n$ integers representing the found cards. Compute their sum, $S$. The goal is now to reduce this sum to zero.
- Take the absolute value of $S$, because the direction of adjustment is irrelevant; we can add positive or negative cards depending on the sign of $S$.
- Divide $|S|$ by $x$ using integer division to count how many full-size cards are needed. If $|S|$ is not divisible by $x$, we need one additional card to cover the remainder.
- Output the total number of cards calculated.
Why it works: by repeatedly taking cards of the maximum allowed absolute value, we minimize the number of cards needed. The invariant is that after each added card, the remaining sum to fix decreases by at most $x$. Using ceiling division ensures that any remaining part smaller than $x$ is counted as one extra card. This guarantees the minimum number.
Python Solution
import sys
input = sys.stdin.readline
n, x = map(int, input().split())
cards = list(map(int, input().split()))
total = sum(cards)
needed_cards = (abs(total) + x - 1) // x # ceiling division
print(needed_cards)
We first read the input and compute the sum of the found cards. The expression (abs(total) + x - 1) // x calculates the ceiling of abs(total) / x efficiently without floating-point operations. This avoids off-by-one errors that occur with naive integer division when the sum is not exactly divisible by $x$.
Worked Examples
Sample 1
Input:
3 2
-1 1 2
| Step | total | abs(total) | cards needed |
|---|---|---|---|
| initial | 2 | 2 | (2 + 2 - 1)//2 = 1 |
Explanation: The sum of the found cards is 2. The maximum single card value is 2, so one card with -2 is sufficient.
Sample 2
Input:
3 3
1 1 1
| Step | total | abs(total) | cards needed |
|---|---|---|---|
| initial | 3 | 3 | (3 + 3 - 1)//3 = 1 |
Here the sum is exactly divisible by the maximum card value, so only one card is needed.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n) | We compute the sum of n cards once. |
| Space | O(n) | We store the list of n integers. |
For the given constraints, n ≤ 1000, this solution runs comfortably within the 1-second limit. Memory usage is also minimal.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
n, x = map(int, input().split())
cards = list(map(int, input().split()))
total = sum(cards)
return str((abs(total) + x - 1) // x)
# provided samples
assert run("3 2\n-1 1 2\n") == "1", "sample 1"
assert run("3 3\n1 1 1\n") == "1", "sample 2"
# custom cases
assert run("1 1\n7\n") == "7", "single card, large sum"
assert run("4 5\n5 5 5 5\n") == "4", "sum divisible by x"
assert run("2 10\n-3 4\n") == "1", "sum small, single card covers"
assert run("5 1\n1 -1 1 -1 1\n") == "1", "alternating sum"
| Test input | Expected output | What it validates |
|---|---|---|
| 1 1\n7 | 7 | Single card far from zero, multiple needed |
| 4 5\n5 5 5 5 | 4 | Sum divisible by max card |
| 2 10\n-3 4 | 1 | Small sum with x larger than remainder |
| 5 1\n1 -1 1 -1 1 | 1 | Alternating values, remainder handling |
Edge Cases
If the sum of cards is zero, the algorithm immediately returns 0 because no new cards are required. For example, input 2 5\n3 -3 gives total = 0, so (abs(0) + 5 - 1) // 5 = 0.
If the absolute sum is smaller than x, the algorithm correctly outputs 1, covering the scenario where a single card suffices. For input 1 5\n3, abs(3)/5 is less than 1, so ceiling division produces 1.
In all other cases, ceiling division ensures we always pick the minimum number of cards necessary without exceeding the required sum.
This approach is direct, handles all edge cases, and scales trivially to the input limits.