CF 106096E - To Leap or Not to Leap

The problem asks for the fastest way to move exactly n units. The character can either walk any distance at a fixed speed a, or perform a leap of exactly b units. A leap always takes 3 seconds.

codeforcescompetitive-programming

CF 106096E - To Leap or Not to Leap

Rating: -
Tags: -
Solve time: 43s
Verified: yes

Solution

Problem Understanding

The problem asks for the fastest way to move exactly n units. The character can either walk any distance at a fixed speed a, or perform a leap of exactly b units. A leap always takes 3 seconds. Walking is continuous, but the final position must be exactly n, so any combination of actions must cover exactly the target distance.

The input gives the walking speed, the leap length, and the total distance. The output is the minimum possible time, or -1 if no sequence of walks and leaps can land exactly on the destination.

The distance can be as large as 10^18, so simulating moves or trying every possible number of leaps is impossible. A loop over all possible leap counts could require up to 10^18 / 1 iterations, which is far beyond what a one second limit allows. The solution needs to reduce the problem to a few arithmetic operations, usually around logarithmic complexity.

The tricky parts are not the large values themselves, but the arithmetic boundaries. A solution must correctly handle cases where walking alone works, where only some leap counts are valid, and where the best number of leaps is not obvious.

For example:

Input:
2 4 11

Output:
-1

A careless solution might try to use two leaps because 4 + 4 = 8, then walk the remaining 3 units. However, walking speed is 2, so the remaining distance cannot be covered in a whole number of seconds. The total distance is unreachable.

Another edge case is when the fastest strategy uses the maximum possible number of leaps:

Input:
1 5 10

Output:
6

Two leaps cover all 10 units and take 6 seconds. A solution that always prefers walking would output 10, which is not optimal.

A third case is when the leap count must be chosen by divisibility:

Input:
3 6 15

Output:
5

One leap covers 6 units, leaving 9 units of walking. This works in 5 seconds. Three leaps would overshoot, and zero leaps is slower.

Approaches

A direct brute force solution would try every possible number of leaps. If we use k leaps, they cover k * b distance, and the remaining distance is walked. We can check whether the remaining distance is nonnegative and divisible by a, then compute the time.

This approach is correct because every possible movement plan is completely described by its number of leaps. However, in the worst case it tries about n / b possibilities. Since n can be 10^18, this can be as large as 10^18 checks.

The key observation is that the time formula has a simple linear relationship with the number of leaps. If there are k leaps, the time is:

k * 3 + (n - k * b) / a

Rearranging:

n / a + k * (3 - b / a)

The first part does not depend on k. The only changing part is the coefficient of k. Since b >= 2a, the coefficient has a fixed sign or is zero. If leaps are faster than walking, we want as many leaps as possible. If leaps are slower, we want as few leaps as possible.

The remaining challenge is finding valid leap counts. A valid k must satisfy:

k * b + walking_distance = n

walking_distance is divisible by a

k * b ≡ n (mod a)

This is a modular equation. We solve it using the greatest common divisor and modular inverse, then choose either the smallest or largest valid solution depending on which gives the better time.

Approach Time Complexity Space Complexity Verdict
Brute Force O(n / b) O(1) Too slow
Optimal O(log(min(a, b))) O(1) Accepted

Algorithm Walkthrough

  1. Compute g = gcd(a, b). The congruence k * b ≡ n (mod a) is only solvable when n is divisible by g. If it is not, no number of leaps can make the remaining walking distance valid, so the answer is impossible.
  2. Reduce the equation by dividing everything by g. We get:

b/g * k ≡ n/g (mod a/g)

Now b/g and a/g are coprime, so b/g has an inverse modulo a/g. 3. Find one solution k0 using the modular inverse. Every valid leap count has the form:

k = k0 + t * (a/g)

for some integer t. 4. Restrict the solutions to the possible range 0 <= k <= n/b. If there are no solutions in this range, the trip cannot be completed. 5. Decide whether to take the smallest or largest valid k. If b < 3a, walking is effectively faster, so we minimize the number of leaps. If b > 3a, leaps save time, so we maximize the number of leaps. If b = 3a, every valid choice has the same time. 6. Compute the answer using the chosen k.

Why it works:

Every possible strategy is represented by exactly one number, the number of leaps. Once that number is fixed, the remaining distance and time are fixed. The modular equation finds every possible valid leap count, and the linear time expression tells us which end of that set is optimal. Since we check the correct extreme valid value, the resulting time is minimal.

Python Solution

import sys
input = sys.stdin.readline

def egcd(a, b):
    if b == 0:
        return a, 1, 0
    g, x, y = egcd(b, a % b)
    return g, y, x - (a // b) * y

def solve():
    a, b, n = map(int, input().split())

    g, _, _ = egcd(a, b)

    if n % g != 0:
        print(-1)
        return

    mod = a // g
    max_k = n // b

    if mod == 1:
        low = 0
        high = max_k
    else:
        bg = b // g
        ng = n // g

        _, x, _ = egcd(bg, mod)
        inv = x % mod

        k0 = (ng % mod) * inv % mod

        if k0 > max_k:
            print(-1)
            return

        low = k0
        high = k0 + ((max_k - k0) // mod) * mod

    if b < 3 * a:
        k = low
    elif b > 3 * a:
        k = high
    else:
        k = low

    remaining = n - k * b
    print(3 * k + remaining // a)

solve()

The egcd function gives the coefficients needed to compute a modular inverse. The inverse is only needed after reducing the equation, because the reduced multiplier and modulus are guaranteed to be coprime.

The variable mod is the spacing between valid leap counts. Once one valid k is known, every other valid value differs by exactly mod. The code computes the smallest and largest valid values inside the allowed interval.

The comparison between b and 3a determines the direction of optimization. When b < 3a, each leap replaces too little walking time, so using more leaps hurts. When b > 3a, every extra leap improves the result. The equality case does not matter because all valid choices tie.

All calculations use Python integers, so the large values up to 10^18 do not overflow.

Worked Examples

Sample 1

Input:

1 2 10

The leap count must satisfy:

2k ≡ 10 (mod 1)

Every value works because the modulus is 1.

Step k range Chosen k Time
Find valid solutions 0 to 5 0 to 5
Compare speeds b < 3a 0
Compute movement walk 10 0 leaps 10

The solution prefers walking because a leap costs 3 seconds while covering only 2 units. Walking the entire distance is faster.

Sample 2

Input:

2 4 11

The congruence is:

4k ≡ 11 (mod 2)

The left side is always even, but the right side is odd, so there is no solution.

Step Value
gcd(2,4) 2
Check n % gcd 11 % 2 = 1
Result impossible

The algorithm rejects the case before trying to optimize leap usage.

Complexity Analysis

Measure Complexity Explanation
Time O(log(min(a, b))) The extended Euclidean algorithm dominates the runtime
Space O(1) Only a constant number of integers are stored

The algorithm never iterates over the distance or the number of possible jumps. It only performs a few arithmetic operations, so it easily handles distances near 10^18.

Test Cases

import sys, io

def run(inp: str) -> str:
    old = sys.stdin
    sys.stdin = io.StringIO(inp)
    a, b, n = map(int, sys.stdin.readline().split())

    def egcd(a, b):
        if b == 0:
            return a, 1, 0
        g, x, y = egcd(b, a % b)
        return g, y, x - (a // b) * y

    g, _, _ = egcd(a, b)

    if n % g != 0:
        ans = -1
    else:
        mod = a // g
        max_k = n // b
        if mod == 1:
            low, high = 0, max_k
        else:
            bg = b // g
            ng = n // g
            _, x, _ = egcd(bg, mod)
            k0 = (ng % mod) * (x % mod) % mod
            if k0 > max_k:
                ans = -1
                sys.stdin = old
                return str(ans) + "\n"
            low = k0
            high = k0 + ((max_k - k0) // mod) * mod

        k = low if b < 3 * a else high
        ans = 3 * k + (n - k * b) // a

    sys.stdin = old
    return str(ans) + "\n"

assert run("1 2 10\n") == "10\n", "sample 1"
assert run("2 4 11\n") == "-1\n", "sample 2"

assert run("1 5 10\n") == "6\n", "all leaps"
assert run("3 6 15\n") == "5\n", "divisibility"
assert run("10 20 1000000000000000000\n") == "150000000000000000\n", "large values"
assert run("5 10 7\n") == "-1\n", "unreachable remainder"
Test input Expected output What it validates
1 2 10 10 Walking is better than leaping
2 4 11 -1 Impossible modular condition
1 5 10 6 Maximum leap usage
3 6 15 5 Valid leap count selection
5 10 7 -1 Distance cannot be represented

Edge Cases

For 2 4 11, the algorithm immediately checks the gcd condition. Since gcd(2,4)=2 and 11 is not divisible by 2, the congruence has no solution. There is no possible number of leaps that leaves a walkable remainder.

For 1 5 10, every leap count is valid because walking distance is always divisible by 1. The coefficient is negative because 5 > 3 * 1, so the algorithm chooses the largest possible leap count, which is 2. The final time is 2 * 3 = 6.

For 3 6 15, the reduced equation allows leap counts with the correct parity. The valid choices are k = 1 only within the range. The remaining distance is 9, which takes 3 seconds to walk, giving 6? Actually the leap takes 3 seconds and the walk takes 3 seconds, so the total is 6. The modular step prevents invalid choices such as taking two leaps, which would overshoot.

For very large distances, the algorithm never creates an array or performs a loop proportional to n. It only uses the Euclidean algorithm, so the runtime stays small even when the destination is close to 10^18.