結果
| 問題 |
No.968 引き算をして門松列（その３）
|
| ユーザー |
 lam6er
|
| 提出日時 | 2025-04-16 16:24:00 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 6,737 bytes |
| コンパイル時間 | 347 ms |
| コンパイル使用メモリ | 81,628 KB |
| 実行使用メモリ | 83,552 KB |
| 最終ジャッジ日時 | 2025-04-16 16:24:42 |
| 合計ジャッジ時間 | 2,359 ms |
|
ジャッジサーバーID （参考情報） |
judge3 / judge4 |
（要ログイン）
| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 |
| other | AC * 1 WA * 9 |
ソースコード
import sys
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 compute_min_cost(A, B, C, X, Y, Z):
if is_kadomatsu(A, B, C):
return 0
min_cost = float('inf')
# Case 1: B is max, A is second, C is min (B > A > C)
# Try using operation Y (B and C)
lower = max(C - A + 1, 0)
upper = min(B - A - 1, C - 1, B - 1)
if lower <= upper:
k = lower
cost = Y * k
new_B = B - k
new_A = A
new_C = C - k
if new_B > new_A and new_A > new_C and new_B > 0 and new_C > 0 and new_A > 0:
if is_kadomatsu(new_A, new_B, new_C):
min_cost = min(min_cost, cost)
# Try using operation Z (A and C)
if A > C:
lower_k = max(A - B + 1, 0)
upper_k = min(A - 1, C - 1)
if lower_k <= upper_k:
k = lower_k
new_A_val = A - k
new_C_val = C - k
new_B_val = B
if new_B_val > new_A_val and new_A_val > new_C_val and new_A_val > 0 and new_C_val > 0 and new_B_val > 0:
if is_kadomatsu(new_A_val, new_B_val, new_C_val):
cost = Z * k
min_cost = min(min_cost, cost)
# Case 2: B is max, C is second, A is min (B > C > A)
# Try using operation Y (B and C)
lower = max(A - C + 1, 0)
upper = min(B - C - 1, A - 1, B - 1)
if lower <= upper:
k = lower
cost = Y * k
new_B = B - k
new_C_val = C - k
new_A_val = A
if new_B > new_C_val and new_C_val > new_A_val and new_B > 0 and new_A_val > 0 and new_C_val > 0:
if is_kadomatsu(new_A_val, new_B, new_C_val):
min_cost = min(min_cost, cost)
# Try using operation X (A and B)
if C > A:
lower_k = max(C - B + 1, 0)
upper_k = min(C - 1, A - 1)
if lower_k <= upper_k:
k = lower_k
new_A_val = A - k
new_B_val = B - k
new_C_val = C
if new_B_val > new_C_val and new_C_val > new_A_val and new_A_val > 0 and new_B_val > 0 and new_C_val > 0:
if is_kadomatsu(new_A_val, new_B_val, new_C_val):
cost = X * k
min_cost = min(min_cost, cost)
# Case 3: B is min, A is max, C is second (A > C > B)
# Try using operation X (A and B)
lower = max(B - C + 1, 0)
upper = min(A - C - 1, B - 1, A - 1)
if lower <= upper:
k = lower
cost = X * k
new_A_val = A - k
new_B_val = B - k
new_C_val = C
if new_A_val > new_C_val and new_C_val > new_B_val and new_A_val > 0 and new_B_val > 0 and new_C_val > 0:
if is_kadomatsu(new_A_val, new_B_val, new_C_val):
min_cost = min(min_cost, cost)
# Try using operation Z (A and C)
if C > B:
lower_k = max(C - A + 1, 0)
upper_k = min(C - 1, B - 1)
if lower_k <= upper_k:
k = lower_k
new_A_val = A - k
new_C_val = C - k
new_B_val = B
if new_A_val > new_C_val and new_C_val > new_B_val and new_A_val > 0 and new_B_val > 0 and new_C_val > 0:
if is_kadomatsu(new_A_val, new_B_val, new_C_val):
cost = Z * k
min_cost = min(min_cost, cost)
# Case 4: B is min, C is max, A is second (C > A > B)
# Try using operation Y (B and C)
lower = max(B - A + 1, 0)
upper = min(C - A - 1, B - 1, C - 1)
if lower <= upper:
k = lower
cost = Y * k
new_B_val = B - k
new_C_val = C - k
new_A_val = A
if new_C_val > new_A_val and new_A_val > new_B_val and new_C_val > 0 and new_A_val > 0 and new_B_val > 0:
if is_kadomatsu(new_A_val, new_B_val, new_C_val):
min_cost = min(min_cost, cost)
# Try using operation Z (A and C)
if A > B:
lower_k = max(A - C + 1, 0)
upper_k = min(A - 1, B - 1)
if lower_k <= upper_k:
k = lower_k
new_A_val = A - k
new_C_val = C - k
new_B_val = B
if new_C_val > new_A_val and new_A_val > new_B_val and new_C_val > 0 and new_A_val > 0 and new_B_val > 0:
if is_kadomatsu(new_A_val, new_B_val, new_C_val):
cost = Z * k
min_cost = min(min_cost, cost)
# Case 5: A is max, C is second, B is mid (A > C > B)
# Try using operation Z (A and C)
lower_k = 0
upper_k = min(A - 1, C - 1)
possible = False
if A > C and C > B:
possible = True
else:
diff_A_C = A - C
diff_C_B = C - B
if diff_A_C >= 0 and diff_C_B >= 0:
upper_k = min(upper_k, diff_A_C, diff_C_B)
possible = True
if possible:
for k in [0, 1]:
if k > upper_k:
continue
new_A_val = A - k
new_C_val = C - k
new_B_val = B
if new_A_val > new_C_val and new_C_val > new_B_val and new_A_val > 0 and new_C_val > 0 and new_B_val > 0:
if is_kadomatsu(new_A_val, new_B_val, new_C_val):
cost = Z * k
min_cost = min(min_cost, cost)
# Case 6: C is max, A is second, B is mid (C > A > B)
# Try using operation Z (A and C)
lower_k = 0
upper_k = min(C - 1, A - 1)
possible = False
if C > A and A > B:
possible = True
else:
diff_C_A = C - A
diff_A_B = A - B
if diff_C_A >= 0 and diff_A_B >= 0:
upper_k = min(upper_k, diff_C_A, diff_A_B)
possible = True
if possible:
for k in [0, 1]:
if k > upper_k:
continue
new_A_val = A - k
new_C_val = C - k
new_B_val = B
if new_C_val > new_A_val and new_A_val > new_B_val and new_C_val > 0 and new_A_val > 0 and new_B_val > 0:
if is_kadomatsu(new_A_val, new_B_val, new_C_val):
cost = Z * k
min_cost = min(min_cost, cost)
# Check other possible combinations
# Additional cases can be added here if necessary
return min_cost if min_cost != float('inf') else -1
def main():
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
print(compute_min_cost(A, B, C, X, Y, Z))
if __name__ == '__main__':
main()