CF 104246H - How Far have You been?

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Solve time: 1m
Verified: yes

Solution

Problem Understanding

We are simulating a single move in a simplified board game. The board has 100 cells arranged in a path, and Farha is currently fixed at cell 94. She rolls a die once, producing an integer k between 1 and 6, and immediately moves forward k steps, landing on cell 94 + k.

After this move, the board may have special rules that change her position. Some cells act like ladders or snakes: landing on a ladder cell changes her state in a positive way (conceptually “she is on a ladder”), and landing on a snake cell moves her down to a lower-numbered cell, after which we must report the new position.

The task is to determine what happens immediately after this single move: whether she wins by reaching exactly 100, whether she lands on a ladder-triggering cell, whether she gets bitten by a snake and is moved elsewhere, or whether nothing special happens.

The input is just the dice roll k, so the entire state transition is deterministic from a fixed starting point. The output is a single message describing the outcome.

Since k is at most 6, there is no algorithmic complexity concern. This is constant-time logic. Any solution that explicitly hardcodes or checks conditions for each possible outcome is sufficient.

The main subtlety is that multiple outcomes depend on the destination cell after the move. A naive mistake would be to treat conditions independently of the landing cell, or to forget that snake effects override the displayed position.

Edge cases are small but important because the board is partially unspecified in the statement text and is implicitly defined by sample behavior. The key is that all special cases are tied to specific landing cells after moving from 94.

For example, if k = 6, she lands on 100 and wins immediately, so no ladder or snake message should appear even if 100 is hypothetically part of another rule set. If k = 5, she lands on 99, which triggers a ladder condition in the samples. If k = 1, she lands on 95 and is bitten by a snake that sends her to 75, so the final output must reflect the post-snake position.

Approaches

A brute-force interpretation would be to explicitly simulate the board game rules cell by cell: move from 94 to 94 + k, then consult a full board representation that encodes whether each cell is a snake, a ladder, or normal, and then apply transitions until no more changes occur. This is the standard approach for a full snakes-and-ladders simulation, where each cell might have chained transitions.

That general simulation would still be correct here, but it is unnecessary because the starting position is fixed and the number of possible moves is extremely small. The worst case for a full simulation would be O(1) per test case anyway, but with a larger board or multiple moves it would scale poorly.

The key observation is that the entire game state collapses into a tiny decision table indexed only by the final position after the dice roll. Since k is at most 6, there are only six possible outcomes, and each one deterministically maps to a single printed response.

So instead of simulating a dynamic system, we reduce the problem to a direct lookup: compute final position p = 94 + k, and then check p against the known special cases.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Simulation	O(1) per move (general case O(steps))	O(1) or O(100)	Accepted but overkill
Direct Case Analysis	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the integer k and compute the landing position p = 94 + k. This represents Farha’s exact cell after the dice roll, and every decision depends only on this value.
Check whether p equals 100. If so, the game ends immediately in a win state. No other rule applies because reaching the final cell overrides all intermediate mechanics.
Check whether p corresponds to a snake-activated cell. In the sample behavior, p = 95 triggers a snake that moves her to 75. In this case, we must output the post-snake position, not the landing cell, because the game state updates after the snake bite.
Check whether p corresponds to a ladder-triggering cell. In the sample behavior, p = 99 triggers a ladder condition. The output does not require a new position, only a status message indicating the ladder event.
If none of the above conditions apply, output that nothing special happens, since the landing cell has no modifiers.

The ordering matters because terminal conditions must be prioritized. Winning at 100 must override any other interpretation. Snake movement must be applied before printing the final position. Ladder detection is a status-only condition that applies after ensuring no snake or win occurred.

Why it works

The process is a deterministic function of a single variable p. Every possible outcome is mutually exclusive and tied to exact cell identities. Since there is no chaining of moves beyond one optional snake jump, the system forms a direct partition of the small domain {95, 96, 97, 98, 99, 100}. Evaluating conditions in a fixed priority order guarantees that each p maps to exactly one valid output, preserving correctness without needing simulation or backtracking.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    k = int(input().strip())
    p = 94 + k

    if p == 100:
        print("Yay! Farha has won the game. She is now at 100.")
    elif p == 95:
        print("Alas! Farha is bitten by snake. She is now at 75.")
    elif p == 99:
        print("Farha is on ladder.")
    else:
        print("Nothing happened to her.")

if __name__ == "__main__":
    solve()

The solution computes the landing cell once and branches based on its value. The ordering ensures that the win condition is checked first, since reaching 100 must not be overridden. The snake case explicitly replaces the position with 75 as required by the sample, which is crucial because the output depends on the post-transition state. The ladder case only prints a message and does not modify position. All remaining cases fall through to the default message.

Worked Examples

Sample 1

Input k = 1 gives p = 95.

Step	k	p	Condition	Output
1	1	95	snake	bitten, move to 75

This demonstrates the snake transition rule where landing on 95 does not end the game but instead forces a backward move. The key invariant is that snake effects must be applied before output.

Sample 2

Input k = 6 gives p = 100.

Step	k	p	Condition	Output
1	6	100	win	game won

This confirms that reaching 100 is a terminal condition that overrides all others, ensuring no further rule evaluation is needed.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only a few integer operations and comparisons are performed
Space	O(1)	No auxiliary data structures are used

The constraints restrict k to a constant range, so the solution is effectively constant-time regardless of input distribution. This comfortably satisfies both time and memory limits.

Test Cases

import sys, io

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

    from contextlib import redirect_stdout
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    builtins.input = input_backup
    return out.getvalue().strip()

def solve():
    k = int(input().strip())
    p = 94 + k

    if p == 100:
        print("Yay! Farha has won the game. She is now at 100.")
    elif p == 95:
        print("Alas! Farha is bitten by snake. She is now at 75.")
    elif p == 99:
        print("Farha is on ladder.")
    else:
        print("Nothing happened to her.")

# provided samples
assert run("1\n") == "Alas! Farha is bitten by snake. She is now at 75.", "sample 1"
assert run("6\n") == "Yay! Farha has won the game. She is now at 100.", "sample 2"
assert run("5\n") == "Farha is on ladder.", "sample 3"

# custom cases
assert run("2\n") == "Nothing happened to her.", "normal move"
assert run("3\n") == "Nothing happened to her.", "normal move"
assert run("4\n") == "Nothing happened to her.", "normal move"
assert run("1\n") == "Alas! Farha is bitten by snake. She is now at 75.", "snake boundary"

Test input	Expected output	What it validates
1	snake to 75	snake transition correctness
6	win message	terminal condition priority
5	ladder message	special non-terminal event
2	nothing	default behavior

Edge Cases

The only meaningful edge cases are the boundary landing cells where special rules activate.

For k = 1, the position becomes 95. The algorithm checks p == 100 first, which fails, then detects the snake condition at 95. The output reflects the updated position 75, confirming that post-move transformations are applied correctly.

For k = 6, the position becomes 100. The algorithm immediately matches the win condition and prints the victory message. No snake or ladder checks are reached, ensuring correct prioritization of terminal state.

For k = 5, the position becomes 99. The snake and win checks fail, and the ladder condition triggers, producing the ladder message. Since ladders do not change position in the output, no further processing is required.