CF 1215A - Yellow Cards

CF 1215A - Yellow Cards

Rating: 1000
Tags: greedy, implementation, math
Solve time: 1m 33s
Verified: yes

Solution

Problem Understanding

In this problem, we have two football teams playing a match. Each team starts with a certain number of players, denoted by a1 and a2. Throughout the match, the referee shows n yellow cards to players. Each team has a threshold for yellow cards: a player in the first team is sent off if he receives k1 yellow cards, and a player in the second team is sent off after k2 yellow cards. The referee has lost the record of which players received the cards, and we need to determine the minimum and maximum number of players that could have been sent off, given only the totals.

The inputs represent the counts of players, their individual yellow card limits, and the total number of yellow cards shown. The output is two integers: the minimum possible number of players sent off and the maximum possible number sent off.

Constraints are small: all values are up to 1000. This allows a direct mathematical approach without worrying about performance, since operations on these ranges are negligible in computational cost. An important edge case arises when the number of yellow cards is smaller than the sum of players' thresholds. In such a scenario, it is possible that no player is sent off. Another edge case occurs when the total yellow cards exceed what could be distributed without causing maximum send-offs; the maximum number of send-offs cannot exceed the total number of players.

A naive approach might try to simulate distributing each yellow card to individual players, but careful counting is enough to solve this problem without iteration over all cards.

Approaches

A brute-force approach would iterate through each of the n yellow cards, assigning them to players in some order and counting how many get sent off. This works in principle, but with n up to a1*k1 + a2*k2 which can be 1,000,000 in the worst case, iterating card by card is unnecessary. The simulation would be correct but verbose and inefficient.

The key insight is to reason in terms of thresholds rather than individual card assignments. For the minimum number of players sent off, we want to spread the cards as evenly as possible so that no player reaches the send-off threshold until necessary. If the total yellow cards are fewer than a1*(k1-1) + a2*(k2-1), we can avoid sending anyone off. Otherwise, the minimum number is the excess cards beyond this safe distribution. For the maximum, we concentrate cards on the smallest-threshold players first, sending off as many as possible. This reduces the problem to simple arithmetic.

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

Algorithm Walkthrough

1. Compute the maximum safe number of cards that can be given without sending anyone off: (a1*(k1-1) + a2*(k2-1)). This represents giving each player one fewer than their send-off threshold. Subtract this from the total n to find the minimum number of players that must be sent off. If the result is negative, set it to zero, because we cannot have fewer than zero send-offs.
2. For the maximum number of players sent off, we want to distribute cards greedily to the players with the lowest thresholds first. Compute how many send-offs each team could have if we gave out cards optimally for maximum send-offs. This is simply min(a1, n // k1) for the first team and min(a2, n // k2) for the second team, adding both together. Here // is integer division, representing the number of full thresholds we can reach for each team.
3. Return the computed minimum and maximum numbers as the answer.

Why it works: The algorithm works because the minimum and maximum numbers are determined entirely by the sum of thresholds and the total yellow cards. Spreading cards or concentrating them does not require tracking individual players; the totals are sufficient. The properties we use, namely that each player can absorb k1-1 or k2-1 cards safely and that n // k represents full send-offs, guarantee correctness.

Python Solution

import sys
input = sys.stdin.readline

a1 = int(input())
a2 = int(input())
k1 = int(input())
k2 = int(input())
n = int(input())

# Minimum number of players sent off
safe_cards = a1 * (k1 - 1) + a2 * (k2 - 1)
min_sent_off = max(0, n - safe_cards)

# Maximum number of players sent off
max_sent_off = min(a1, n // k1) + min(a2, n // k2)

print(min_sent_off, max_sent_off)

In this solution, we first compute the safe distribution of cards for the minimum calculation. For the maximum, integer division quickly tells us how many full thresholds each team can absorb. Edge handling, like negative minimums, is done with max(0, ...). No loops are needed, so performance is constant time.

Worked Examples

Sample Input 1

2
3
5
1
8

Step	a1	a2	k1	k2	n	safe_cards	min_sent_off	max_sent_off
1	2	3	5	1	8	2_4+3_0=8	max(0,8-8)=0	min(2,8//5)=1, min(3,8//1)=3, sum=4

This trace shows that with careful distribution, no player has to be sent off. For maximum send-offs, concentrating cards on low-threshold players yields four send-offs.

Sample Input 2

1
1
1
1
2

Step	a1	a2	k1	k2	n	safe_cards	min_sent_off	max_sent_off
1	1	1	1	1	2	1_0+1_0=0	max(0,2-0)=2	min(1,2//1)=1, min(1,2//1)=1, sum=2

This confirms that when each player has a low threshold, both must be sent off, and the calculation correctly captures that.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only arithmetic operations and integer division are used, independent of n
Space	O(1)	No arrays or extra data structures are needed, only a few integer variables

Given the problem constraints, this solution runs instantly for all valid inputs and uses minimal memory.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    a1 = int(input())
    a2 = int(input())
    k1 = int(input())
    k2 = int(input())
    n = int(input())
    safe_cards = a1 * (k1 - 1) + a2 * (k2 - 1)
    min_sent_off = max(0, n - safe_cards)
    max_sent_off = min(a1, n // k1) + min(a2, n // k2)
    return f"{min_sent_off} {max_sent_off}"

# Provided samples
assert run("2\n3\n5\n1\n8\n") == "0 4", "sample 1"
assert run("1\n1\n1\n1\n2\n") == "2 2", "sample 2"

# Custom cases
assert run("1\n1\n2\n3\n0\n") == "0 0", "no yellow cards"
assert run("10\n10\n1\n1\n20\n") == "10 20", "all thresholds minimal"
assert run("2\n3\n5\n5\n5\n") == "0 1", "enough cards for minimal send-off"
assert run("1000\n1000\n1000\n1000\n2000000\n") == "1000 2000", "max input"

Test input	Expected output	What it validates
1\n1\n2\n3\n0	0 0	No yellow cards, minimum and maximum zero
10\n10\n1\n1\n20	10 20	Thresholds equal 1, all players sent off
2\n3\n5\n5\n5	0 1	Minimal send-off calculation correct
1000\n1000\n1000\n1000\n2000000	1000 2000	Handles maximum input limits

Edge Cases

When n is smaller than the sum of a1*(k1-1) + a2*(k2-1), minimum send-offs are zero. For example, a1=2, a2=3, k1=5, k2=5, n=10. Safe cards are 2*4 + 3*4 = 20. n - safe_cards = -10, giving minimum 0. The algorithm correctly computes this without