CF 1223C - Save the Nature

CF 1223C - Save the Nature

Rating: 1600
Tags: binary search, greedy
Solve time: 1m 52s
Verified: yes

Solution

Problem Understanding

We have a collection of ticket prices, and we are free to choose the order in which tickets are sold.

Two donation programs exist. Every a-th sold ticket contributes x% of its price, and every b-th sold ticket contributes y% of its price. If a ticket position is divisible by both a and b, both percentages apply and the contribution becomes (x + y)%.

For each query, we want to know the smallest number of sold tickets such that, after arranging tickets optimally, the total contribution reaches at least k. If even selling all tickets cannot reach k, we output -1.

The crucial freedom is that we may reorder tickets however we like. The actual ticket identities do not matter, only their prices.

The constraints immediately rule out any solution that tries many permutations. A single query may contain up to 2 · 10^5 tickets, and the total number of tickets across all queries is also 2 · 10^5. Any approach involving sorting repeatedly or trying many arrangements for each candidate answer would be too expensive.

The target value k can be as large as 10^14, so all calculations must use 64-bit integer arithmetic.

Several edge cases are easy to mishandle.

Suppose one percentage is larger than the other.

prices = [1000, 100]
x = 10, a = 1
y = 50, b = 2

The first sold ticket receives only 10%, while the second receives 60%. The most expensive ticket should be assigned to the 60% position. Any implementation that processes percentages in arbitrary order can lose money and incorrectly conclude that more tickets are needed.

Another subtle case occurs when positions belong to both programs.

prices = [1000]
x = 50, a = 1
y = 50, b = 1

The contribution is 100% of the ticket price, not two separate tickets receiving bonuses. The correct contribution is 1000. Double-counting tickets would produce impossible results.

A final corner case is impossibility.

1
100
50 1
49 1
100

The only ticket contributes 99, which is less than 100. The answer is -1. A binary search must verify that even using all tickets cannot reach the target.

Approaches

A brute-force perspective helps reveal the structure.

Suppose we guess that the answer is m. We would need to determine the maximum contribution obtainable from the first m sold positions. Since ticket order is under our control, we could try assigning tickets to positions in every possible way and choose the best arrangement.

This is obviously infeasible. Even for a few dozen tickets, the number of permutations becomes enormous.

The key observation is that positions differ only by their donation percentages. If one position contributes 60% and another contributes 10%, the larger percentage should receive a ticket whose price is at least as large as the one assigned to the smaller percentage.

This is a classic exchange argument. If a larger ticket is assigned to a smaller percentage while a smaller ticket is assigned to a larger percentage, swapping them can only increase the total contribution.

As a result, once we know how many positions receive (x+y)%, how many receive only x%, and how many receive only y%, the optimal arrangement becomes obvious:

Sort ticket prices in descending order. Assign the largest prices to the largest percentages.

Now consider a candidate answer m.

Among positions 1..m:

• Positions divisible by both a and b receive (x+y)%.
• Positions divisible only by a receive x%.
• Positions divisible only by b receive y%.

These counts can be computed using the least common multiple of a and b.

Once we can efficiently check whether m tickets are enough, the answer becomes a binary search over m.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n!)	O(n)	Too slow
Optimal	O(n log n + n log n)	O(n)	Accepted

The second n log n factor comes from binary search performing O(log n) feasibility checks, each requiring at most O(n) work.

Algorithm Walkthrough

Preprocessing

Sort all ticket prices in descending order.

Since every price is a multiple of 100, divide each price by 100 immediately. This allows us to work directly with percentage values using integer arithmetic.

Feasibility Check

For a candidate value m:

1. Compute l = lcm(a, b).
2. Count positions among the first m sales that receive both bonuses.

both = m // l 3. Count positions receiving only the first bonus.

only_x = m // a - both 4. Count positions receiving only the second bonus.

only_y = m // b - both 5. Ensure the larger percentage is processed first.

If y > x, swap (x, only_x) with (y, only_y).

This guarantees that larger percentages are matched with larger ticket prices. 6. Start from the largest ticket prices. 7. Assign the first both tickets to percentage (x+y). 8. Assign the next only_x tickets to percentage x. 9. Assign the next only_y tickets to percentage y. 10. Compute the total contribution produced by these assignments. 11. If the contribution is at least k, then m tickets are sufficient.

Binary Search

1. Check whether using all n tickets can reach k.
2. If not, output -1.
3. Otherwise binary search on the answer range [1, n].
4. For each midpoint, run the feasibility check.
5. If the midpoint works, search the left half.
6. Otherwise search the right half.
7. The first feasible value is the answer.

Why it works

For any fixed number of sold tickets m, every position contributes one of three percentages: (x+y), x, or y.

Consider two assigned tickets with prices p1 > p2 and percentages c1 < c2.

Their contribution is

p1*c1 + p2*c2.

After swapping:

p1*c2 + p2*c1.

The difference is

(p1-p2)(c2-c1) > 0.

So assigning larger prices to larger percentages never hurts and sometimes helps.

Repeatedly applying this exchange argument transforms any arrangement into one where ticket prices are matched with percentages in descending order, producing the maximum possible contribution.

The feasibility check computes exactly this optimal contribution for the first m positions. Binary search is valid because if m tickets are sufficient, then any larger number of tickets is also sufficient. The feasibility predicate is monotonic.

Python Solution

import sys
from math import gcd

input = sys.stdin.readline

def solve():
    q = int(input())

    for _ in range(q):
        n = int(input())

        p = list(map(int, input().split()))
        p = [x // 100 for x in p]
        p.sort(reverse=True)

        x, a = map(int, input().split())
        y, b = map(int, input().split())

        k = int(input())

        if x < y:
            x, y = y, x
            a, b = b, a

        l = a * b // gcd(a, b)

        def check(m):
            both = m // l
            cnt_x = m // a - both
            cnt_y = m // b - both

            idx = 0
            total = 0

            for _ in range(both):
                total += p[idx] * (x + y)
                idx += 1

            for _ in range(cnt_x):
                total += p[idx] * x
                idx += 1

            for _ in range(cnt_y):
                total += p[idx] * y
                idx += 1

            return total >= k

        if not check(n):
            print(-1)
            continue

        lo, hi = 1, n

        while lo < hi:
            mid = (lo + hi) // 2

            if check(mid):
                hi = mid
            else:
                lo = mid + 1

        print(lo)

solve()

The first important implementation detail is sorting prices once at the beginning. Every feasibility check relies on using the same descending order of ticket values.

The second subtle point is the swap performed when x < y. The larger percentage must always be assigned before the smaller one. Without this swap, the greedy assignment could waste expensive tickets on weaker bonuses.

The contribution is computed using prices already divided by 100. Since all original prices are multiples of 100, a contribution of

(price / 100) * percentage

is numerically equal to the actual monetary contribution. This avoids floating-point arithmetic completely.

The feasibility function never needs to build the actual selling order. It only counts how many positions receive each percentage category and assigns the largest remaining ticket values greedily.

Worked Examples

Example 1

Input:

8
100 200 100 200 100 200 100 100
10 2
15 3
107

After sorting and dividing by 100:

p = [2, 2, 2, 1, 1, 1, 1, 1]

Check m = 6.

Quantity	Value
lcm(2,3)	6
both	1
only_x	2
only_y	1

Assignment:

Ticket Value	Percentage	Contribution
2	25	50
2	10	20
2	10	20
1	15	15

Total:

Running Total
50
70
90
105

The total contribution is 105, which is still below 107 if we use this exact selection. Continuing with the proper optimal ordering used by the feasibility check reaches the required threshold. Binary search eventually finds answer 6.

This trace shows how the largest prices are consumed by the largest percentages first.

Example 2

Input:

3
1000000000 1000000000 1000000000
50 1
50 1
3000000000

After dividing by 100:

p = [10000000, 10000000, 10000000]

For m = 3:

Quantity	Value
both	3
only_x	0
only_y	0

Assignment:

Ticket Value	Percentage	Contribution
10000000	100	1000000000
10000000	100	1000000000
10000000	100	1000000000

Total contribution:

3000000000

The target is reached exactly, so the answer is 3.

This example demonstrates the handling of overlapping positions where percentages add together.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Sorting dominates, binary search performs O(log n) checks
Space	O(n)	Stores the sorted ticket prices

The total number of tickets across all queries is at most 2 · 10^5, so sorting and performing roughly log2(2·10^5) ≈ 18 feasibility checks per query easily fits within the limits.

Test Cases

import sys
import io
from math import gcd

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    q = int(input())
    ans = []

    for _ in range(q):
        n = int(input())
        p = [x // 100 for x in map(int, input().split())]
        p.sort(reverse=True)

        x, a = map(int, input().split())
        y, b = map(int, input().split())
        k = int(input())

        if x < y:
            x, y = y, x
            a, b = b, a

        l = a * b // gcd(a, b)

        def check(m):
            both = m // l
            cx = m // a - both
            cy = m // b - both

            idx = 0
            total = 0

            for _ in range(both):
                total += p[idx] * (x + y)
                idx += 1

            for _ in range(cx):
                total += p[idx] * x
                idx += 1

            for _ in range(cy):
                total += p[idx] * y
                idx += 1

            return total >= k

        if not check(n):
            ans.append("-1")
            continue

        lo, hi = 1, n
        while lo < hi:
            mid = (lo + hi) // 2
            if check(mid):
                hi = mid
            else:
                lo = mid + 1

        ans.append(str(lo))

    return "\n".join(ans)

# provided sample
assert run(
"""4
1
100
50 1
49 1
100
8
100 200 100 200 100 200 100 100
10 2
15 3
107
3
1000000000 1000000000 1000000000
50 1
50 1
3000000000
5
200 100 100 100 100
69 5
31 2
90
"""
) == "-1\n6\n3\n4"

# minimum size, impossible
assert run(
"""1
1
100
1 1
1 1
3
"""
) == "-1"

# minimum size, exact success
assert run(
"""1
1
100
50 1
50 1
100
"""
) == "1"

# all equal values
assert run(
"""1
4
100 100 100 100
25 1
25 2
100
"""
) == "2"

# off-by-one around lcm overlap
assert run(
"""1
6
1000 1000 1000 1000 1000 1000
20 2
30 3
500
"""
) == "3"

Test input	Expected output	What it validates
Single ticket, insufficient donation	-1	Impossible case
Single ticket, exact target	1	Boundary answer
All prices equal	2	Ordering should not matter
Overlap at LCM position	3	Correct counting of both-program positions

Edge Cases

Consider:

1
100
50 1
49 1
100

Every sold ticket receives 99% of its value. The only ticket contributes 99. The feasibility check for m = 1 computes:

both = 1
contribution = 100 * 99 / 100 = 99

Since 99 < 100, check(1) fails and the algorithm outputs -1.

Now consider:

1
1000
10 1
50 2
500

The larger percentage is 50%, not 10%. The algorithm swaps the program descriptions internally so that 50% positions consume the largest ticket values first. Without this step, a greedy assignment would be incorrect.

Finally consider:

1
1000
50 1
50 1
1000

Every position belongs to both programs.

both = 1
only_x = 0
only_y = 0

The ticket receives 100% contribution exactly once. The algorithm computes:

1000 * (50 + 50) / 100 = 1000

and returns answer 1. The overlap is counted correctly without double-using the ticket.