結果

問題 No.968 引き算をして門松列（その３）
コンテスト
ユーザー gew1fw
提出日時 2025-06-12 18:35:18
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 3,612 bytes
コンパイル時間 149 ms
コンパイル使用メモリ 82,380 KB
実行使用メモリ 83,680 KB
最終ジャッジ日時 2025-06-12 18:35:43
合計ジャッジ時間 2,163 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other WA * 10
権限があれば一括ダウンロードができます

ソースコード

diff # 

def is_kadomatsu(A, B, C):
    if A == B or B == C or A == C:
        return False
    sorted_vals = sorted([A, B, C])
    second = sorted_vals[1]
    return second == A or second == C

def solve():
    import sys
    input = sys.stdin.read().split()
    idx = 0
    T = int(input[idx])
    idx += 1
    for _ in range(T):
        A = int(input[idx])
        B = int(input[idx+1])
        C = int(input[idx+2])
        X = int(input[idx+3])
        Y = int(input[idx+4])
        Z = int(input[idx+5])
        idx += 6
        
        if is_kadomatsu(A, B, C):
            print(0)
            continue
        
        min_cost = float('inf')
        
        # Try making B the maximum with different operations
        # Operation 2 (Y) once
        new_B = B - 1
        new_A = A
        new_C = C - 1
        if new_B > 0 and new_A > 0 and new_C > 0:
            if new_B > new_A and new_B > new_C and new_A != new_C:
                if is_kadomatsu(new_A, new_B, new_C):
                    cost = Y
                    min_cost = min(min_cost, cost)
        # Operation 1 (X) once
        new_B = B - 1
        new_A = A - 1
        new_C = C
        if new_B > 0 and new_A > 0 and new_C > 0:
            if new_B > new_A and new_B > new_C and new_A != new_C:
                if is_kadomatsu(new_A, new_B, new_C):
                    cost = X
                    min_cost = min(min_cost, cost)
        # Operation 3 (Z) once (only if A != C)
        if A != C:
            new_A = A - 1
            new_C = C - 1
            new_B = B
            if new_A > 0 and new_C > 0 and new_B > 0:
                if new_B > new_A and new_B > new_C and new_A != new_C:
                    if is_kadomatsu(new_A, new_B, new_C):
                        cost = Z
                        min_cost = min(min_cost, cost)
        
        # Try making B the minimum with different operations
        # Operation 1 (X) once
        new_B = B - 1
        new_A = A - 1
        new_C = C
        if new_B > 0 and new_A > 0 and new_C > 0:
            if new_B < new_A and new_B < new_C and new_A != new_C:
                if is_kadomatsu(new_A, new_B, new_C):
                    cost = X
                    min_cost = min(min_cost, cost)
        # Operation 2 (Y) once
        new_B = B - 1
        new_A = A
        new_C = C - 1
        if new_B > 0 and new_A > 0 and new_C > 0:
            if new_B < new_A and new_B < new_C and new_A != new_C:
                if is_kadomatsu(new_A, new_B, new_C):
                    cost = Y
                    min_cost = min(min_cost, cost)
        # Operation 3 (Z) once (only if A != C)
        if A != C:
            new_A = A - 1
            new_C = C - 1
            new_B = B
            if new_A > 0 and new_C > 0 and new_B > 0:
                if new_B < new_A and new_B < new_C and new_A != new_C:
                    if is_kadomatsu(new_A, new_B, new_C):
                        cost = Z
                        min_cost = min(min_cost, cost)
        
        # Try other combinations, like using operation 3 once
        new_A = A - 1
        new_C = C - 1
        new_B = B - 0
        if new_A > 0 and new_C > 0 and new_B > 0:
            if (new_B > new_A and new_B > new_C) or (new_B < new_A and new_B < new_C):
                if new_A != new_C and is_kadomatsu(new_A, new_B, new_C):
                    cost = Z * 1
                    min_cost = min(min_cost, cost)
        
        # Check if any of the operations worked
        if min_cost != float('inf'):
            print(min_cost)
        else:
            print(-1)

if __name__ == "__main__":
    solve()
0