Fight Against Monsters(贪心)

这篇博客讲述了英雄Huriyyah与怪物战斗的故事,分析了一种最小化英雄受到总伤害的贪心策略。问题涉及到计算在每回合中先承受怪物攻击再反击的情况,通过计算每个怪物的性价比,按顺序击败它们以达到最小伤害的目标。

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It is my great honour to introduce myself to you here. My name is Aloysius Benjy Cobweb Dartagnan Egbert Felix Gaspar Humbert Ignatius Jayden Kasper Leroy Maximilian. As a storyteller, today I decide to tell you and others a story about the hero Huriyyah, and the monsters.

Once upon a time, citizens in the city were suffering from ?n powerful monsters. They ate small children who went out alone and even killed innocent persons. Before the hero appeared, the apprehension had overwhelmed the people for several decades.

For the good of these unfortunate citizens, Huriyyah set off to the forest which was the main lair of monsters and fought with ?n fierce and cruel monsters. The health point of the ?i-th monster was ???, and its attack value was ????.

They fought in a cave through a turn-based battle. During each second, the hero Huriyyah was attacked by monsters at first, and the damage was the sum of attack values of all alive

### 关于与'Monsters'相关的二分概念或应用 在处理涉及‘Monsters’的游戏开发或其他应用场景时,二分查找算法能够被用于优化多种操作效率。例如,在游戏中怪物(Monsters)可能具有不同的属性值如攻击力、防御力或者速度等,这些数值通常存储在一个有序数组中。 当需要快速定位特定攻击力量级的Monster实例位置而不必遍历整个列表时,就可以利用二分查找来实现高效检索[^1]。下面给出一段Python代码演示如何基于给定条件筛选符合条件的第一个Monster对象: ```python def binary_search_monster(monsters, target_attack): low, high = 0, len(monsters) - 1 while low <= high: mid = (low + high) // 2 if monsters[mid].attack >= target_attack and (mid == 0 or monsters[mid-1].attack < target_attack): return mid elif monsters[mid].attack < target_attack: low = mid + 1 else: high = mid - 1 return -1 ``` 此函数接收一个按照`attack`属性升序排列好的Monster对象列表以及目标最小攻击力作为参数,并返回第一个不低于指定攻击力水平的对象索引;如果没有找到,则返回-1表示不存在这样的元素[^2]。 对于数据结构方面,考虑到游戏场景下可能会频繁增删节点的情况,除了基本线性表外还可以考虑采用平衡树(Balanced Tree)或是跳跃表(Skip List)这类支持动态调整的数据结构来维护已知Monster集合并保持其内部顺序不变以便随时调用上述二分逻辑进行查询操作[^3]。 为了更好地理解具体的应用方式,请参考如下几个相关问题进一步探讨该主题的不同侧面:
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