CF 102416C - Quick coffee

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Solve time: 2m 32s
Verified: yes

Solution

Problem Understanding

We need to make exactly d dollars of change. The available coin denominations are every integer from a through some upper bound b, inclusive, and we may use any denomination any number of times. The task is to find the smallest possible b for which d can be represented as a sum of these coins.

The input contains a and d, with 1 <= a <= d <= 10^9. Since the values can reach one billion, an algorithm that checks a linear number of possible values is far too expensive for the 1 second limit. We need to reduce the problem to constant-time arithmetic rather than trying candidate values one by one.

The first boundary case is a = d. The only way to make d is to use the coin of value d, so the answer must be d. For example, 10 10 gives 10. A careless approach that assumes at least two coins are needed could incorrectly return a smaller value.

Another small case is a = 1. For example, 1 7 has answer 1, because the coin of value 1 alone can make every positive amount. A search that starts from a + 1 would miss this immediately.

The case where d is divisible by a also matters. For 6 18, three coins of value 6 already make the target, so b = 6. A solution based on rounding d / floor(d / a) incorrectly with floating-point arithmetic could introduce precision problems at large values, so integer ceiling division is preferable.

Approaches

A direct approach is to try every possible upper denomination b, starting at a. For each candidate, we could determine whether d can be formed using coins from a through b. This is correct because we inspect the candidates in increasing order, so the first feasible one is necessarily the minimum.

The problem is that b can be as large as d, and d can be 10^9. In the worst case this means checking roughly 10^9 candidates. Even if each candidate could be tested in constant time, that is about one billion operations, which is nowhere near appropriate for a 1 second time limit.

The useful observation comes from fixing the number of coins rather than fixing the largest denomination. Suppose we use exactly k coins. The smallest possible sum is ka, because every coin is at least a. The largest possible sum is kb, because every coin is at most b.

More importantly, because every denomination between a and b exists, every integer sum between ka and kb is achievable. Thus d can be formed with exactly k coins precisely when

ka <= d <= kb.

The first inequality gives k <= floor(d / a). Let

q = floor(d / a).

If we can use at most q coins, then choosing q coins gives us the largest possible flexibility on the lower bound, because qa <= d. We only need b large enough that q coins can reach d, which means

qb >= d.

Therefore the smallest possible b is

ceil(d / q),

where q = floor(d / a).

There is no reason to consider fewer than q coins. If k < q, then ceil(d / k) >= ceil(d / q), so requiring fewer coins can only make the necessary maximum denomination larger or equal. The brute-force search over possible b has consequently been reduced to two integer divisions.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(d - a + 1)	O(1)	Too slow
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read a and d. These are the minimum available coin value and the target amount.
Compute q = d // a. This is the maximum number of coins of value at least a that can possibly fit into a sum of d.
Compute the smallest denomination bound that lets q coins reach d. We need q * b >= d, so use integer ceiling division:

b = (d + q - 1) // q. 4. Print b. Since q is the largest feasible number of coins, this value is the smallest possible upper denomination.

Why it works

Let q = floor(d / a). Since qa <= d, using q coins does not exceed the target when all of them have value a. We need those q coins to be capable of reaching d, which requires qb >= d.

Because the available denominations form the complete integer interval from a to b, every sum between qa and qb is achievable by exactly q coins. Thus b = ceil(d / q) is sufficient.

For any smaller number of coins k < q, reaching d would require b >= ceil(d / k). Since k < q, we have ceil(d / k) >= ceil(d / q). Hence no smaller b can work with fewer coins. The computed value is both feasible and minimal.

Python Solution

import sys
input = sys.stdin.readline

a, d = map(int, input().split())

q = d // a
b = (d + q - 1) // q

print(b)

The value q is computed with integer division, so it is exactly floor(d / a). Since a <= d, q is always at least 1, which makes the subsequent division safe.

The expression (d + q - 1) // q performs ceiling division without floating-point arithmetic. This matters because the input can contain values up to 10^9, and integer arithmetic gives an exact result with no rounding concerns.

Python integers also have arbitrary precision, so there is no integer overflow issue. In languages with fixed-width integer types, the same formula is still safe for these constraints with standard 64-bit integers.

Worked Examples

For the first sample, a = 19 and d = 60.

Variable	Value	Reason
`a`	19	Minimum coin
`d`	60	Target amount
`q = d // a`	3	At most three coins can be used
`b = ceil(d / q)`	20	Three coins of maximum value 20 can reach 60
Answer	20	Minimum valid upper denomination

With coins 19 and 20, the target is simply 20 + 20 + 20 = 60. With only denomination 19, the possible sums around the target do not include 60, because three coins give 57 and four coins give 76. Thus 20 is exactly the first feasible upper bound.

For the second sample, a = 100 and d = 914.

Variable	Value	Reason
`a`	100	Minimum coin
`d`	914	Target amount
`q = d // a`	9	At most nine coins can be used
`b = ceil(d / q)`	102	Nine coins need maximum denomination at least 102
Answer	102	Minimum valid upper denomination

With denominations from 100 through 102, nine coins can make 914. Start with nine 100 coins for 900, then increase the value of seven of those coins by 2, giving 102 + 102 + 102 + 102 + 102 + 102 + 102 + 100 + 100 = 914. With b = 101, nine coins have maximum total value 909, so that bound is insufficient.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	The algorithm performs a constant number of integer operations.
Space	O(1)	Only a few integer variables are stored.

The constraints allow values up to 10^9, which makes any search proportional to the target potentially reach one billion iterations. The constant-time solution avoids that entirely and easily fits within the 1 second and 256 MB limits.

Test Cases

import sys
import io

def solve(data: str) -> str:
    a, d = map(int, data.split())

    q = d // a
    b = (d + q - 1) // q

    return str(b)

def run(inp: str) -> str:
    return solve(inp).strip()

# Provided samples
assert run("19 60") == "20", "sample 1"
assert run("100 914") == "102", "sample 2"

# Minimum-size input
assert run("1 1") == "1", "minimum values"

# Same minimum and target
assert run("10 10") == "10", "a equals d"

# Divisible case
assert run("6 18") == "6", "d is divisible by a"

# Small boundary case
assert run("3 10") == "5", "ceil(10 / floor(10 / 3)) = 5"

# Large values
assert run("500000000 1000000000") == "500000000", "large divisible case"

Test input	Expected output	What it validates
`1 1`	`1`	Minimum possible input and `q = 1`
`10 10`	`10`	Boundary where the minimum coin equals the target
`6 18`	`6`	Exact divisibility and ceiling division
`3 10`	`5`	Non-divisible case that catches rounding errors
`500000000 1000000000`	`500000000`	Large values and constant-time arithmetic

Edge Cases

For a = d, consider 10 10. The calculation gives q = 10 // 10 = 1, followed by b = ceil(10 / 1) = 10. The result is correct because the only available denomination that can make the target is 10 itself. The algorithm does not accidentally assume that multiple coins are required.

For a = 1, consider 1 7. We obtain q = 7, so b = ceil(7 / 7) = 1. This says that the single denomination 1 is enough. Indeed, seven coins of value 1 make the target exactly. This case also confirms that the answer can equal the smallest possible input value a.

For an exactly divisible target, consider 6 18. Here q = 18 // 6 = 3, and b = ceil(18 / 3) = 6. Three coins of value 6 make 18, so no larger denomination is necessary. The calculation lands exactly on a, which is the smallest possible answer.

For a non-divisible target, consider 3 10. We have q = 10 // 3 = 3. Three coins must reach 10, so the maximum denomination must satisfy 3b >= 10, giving b = 4. The actual formula yields (10 + 3 - 1) // 3 = 12 // 3 = 4, not 3. With denominations 3 and 4, the target is 3 + 3 + 4 = 10. This is the kind of case that exposes an incorrect use of ordinary floor division.

Finally, consider the maximum target a = 500000000, d = 1000000000. We get q = 2, and b = ceil(1000000000 / 2) = 500000000. Two coins of value 500000000 make the target exactly. The computation remains constant-time even at the largest input scale.