6.4（学号：3025）

initial_costs = [2.5, 2.6, 2.8, 3.1]
salvage_values = [2.0, 1.6, 1.3, 1.1]
maintenance_costs = [0.3, 0.8, 1.5, 2.0]

dp = [[float('inf')] * 2 for _ in range(4)]
dp[0][1] = initial_costs[0] + maintenance_costs[0]

for i in range(1, 4):
dp[i][1] = min(dp[i-1][1] + maintenance_costs[i],
initial_costs[i] + maintenance_costs[i])

if i > 0:  
    dp[i][0] = dp[i-1][1] + salvage_values[i-1] 

min_cost = min(dp[3][1],
min(dp[i][0] for i in range(3)))

print(f"最优更新策略下的4年内最小总费用为：{min_cost}万元")

print("学号：3025")