結果

問題 No.531 エヌスクミ島の平和協定
ユーザー qwewe
提出日時 2025-05-14 12:58:10
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 4,458 bytes
コンパイル時間 403 ms
コンパイル使用メモリ 68,236 KB
実行使用メモリ 7,844 KB
最終ジャッジ日時 2025-05-14 12:59:26
合計ジャッジ時間 1,531 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge3
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ファイルパターン 結果
sample AC * 2
other AC * 37
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ソースコード

diff # 

#include <iostream>

int main() {
    // Optimize standard I/O operations for performance.
    // This helps especially with large inputs, though for this problem N <= 100000 isn't huge.
    // It ensures that cin/cout operations are faster.
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);
    
    // Declare variables to store the number of animals (n) and boat capacity (m).
    // Use long long to accommodate n up to 100000.
    long long n; 
    long long m; 
    
    // Read the input values for n and m from standard input.
    std::cin >> n >> m;
    
    // The core logic depends on comparing boat capacity 'm' with the number of animals 'n'.
    
    // Case 1: Boat capacity is sufficient to carry all animals at once.
    if (m >= n) {
        // If m >= n, we can load all n animals onto a single boat and move them
        // from Enuskumi Island to New Enuskumi Island in one trip.
        // This trip takes exactly 1 day.
        std::cout << 1 << std::endl;
    } else {
        // Case 2: Boat capacity is less than the number of animals (m < n).
        // In this case, we cannot move all animals in a single trip. We need to consider
        // if intermediate steps are possible while respecting the rule.
        
        // The rule "every animal must be both a predator and a prey" effectively means that
        // on any island, if there are any animals present, the set of animals must satisfy
        // certain properties related to the cyclic predation structure.
        // Analysis shows that the only "valid" sets of animals on an island are:
        // 1. The empty set (no animals).
        // 2. The set of all N animals.
        // 3. If N is even, the set of all odd-indexed animals (S_odd) and the set of all
        //    even-indexed animals (S_even) are also valid. Both S_odd and S_even have size N/2.
        
        // Any state transition (move) must result in a state where the animal sets on both islands are valid.
        // An intermediate state typically involves splitting the N animals into two non-empty sets:
        // S animals on New Enuskumi and A\S animals remaining on Enuskumi.
        
        // Check if n is odd.
        if (n % 2 != 0) {
            // If n is odd, the only valid sets are the empty set and the set of all N animals.
            // It's impossible to partition the N animals into two non-empty valid sets S and A\S.
            // Since m < n, we cannot move all animals at once. Any partial move (moving a proper subset S)
            // would result in an invalid state because neither S nor A\S would be A or empty.
            // Therefore, the move is impossible if n is odd and m < n.
            std::cout << -1 << std::endl;
        } else {
            // If n is even:
            // The sets S_odd and S_even (each size N/2) provide a valid partition of A.
            // This means an intermediate state where S_odd is on one island and S_even on the other is potentially valid.
            // To reach this state or move from it, we need to transport groups of size N/2.
            
            // Check if the boat capacity m is sufficient to carry a group of size N/2.
            if (m >= n / 2) {
                // If m >= n/2, we have enough capacity to move S_odd or S_even.
                // The move can be done in two steps:
                // Day 1: Move S_odd (for example) from Enuskumi to New Enuskumi. Requires capacity N/2.
                //        After Day 1, state is (S_even on Enuskumi, S_odd on New Enuskumi). This is a valid state.
                // Day 2: Move S_even from Enuskumi to New Enuskumi. Requires capacity N/2.
                //        After Day 2, state is (Empty on Enuskumi, A on New Enuskumi). Goal reached.
                // Total time required is 2 days.
                std::cout << 2 << std::endl;
            } else {
                // If m < n/2:
                // The boat capacity is insufficient to move S_odd or S_even (the smallest non-empty valid proper subsets).
                // Since moving any smaller group S would result in an invalid state (as S and A\S wouldn't both be valid),
                // no valid move is possible from the initial state if m < n/2.
                // Therefore, the move is impossible.
                std::cout << -1 << std::endl;
            }
        }
    }
    
    // Return 0 to indicate successful program execution.
    return 0;
}
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