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3856: Monster

A monster has invaded this kingdom, and Teacher Mai wants to kill it.Monster initially has h HP. And it will die if HP is less than 1.Teacher Mai and monster take turns to do their action. In one round, Teacher Mai can attack the monster so that the HP of the monster will be reduced by a. At the end of this round, the HP of monster will be increased by b.After k consecutive rounds attack, However, he can also choose to take a rest in any round.Output YES if Teacher Mai can kill this monster

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DumpMem and Monster - Virtual Memory Explorers on Windows MobileCE

Windows Mobile Virtual Memory Monster    基于上面这个DumpMem,Windows Mobile Virtual Memory Monster给出了更加形象的图形化表示 参考文章:用于 Pocket PC 2002 上查看虚拟地址空间的 Dumpmem 实用工具Slaying the Virtual Memory Monster - Part I Slaying the Virtual Memory Monster - Part II Visualizing the Windows Mobile Virtual Memory Monster

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    Go 每日一库之 wire

    我们用下面的结构表示,同时编写一个创建方法:type Monster struct { Name string} func NewMonster() Monster { return Monster{Name Monster} func NewMission(p Player, m Monster) Mission { return Mission{p, m}} func (m Mission) Start = nil { return Mission{}, err } monster := NewMonster() mission := NewMission(player, monster) return 假设,我们有关勇士和怪兽的故事有两个结局:type EndingA struct { Player Player Monster Monster} func NewEndingA(p Player, m 上面我们需要注入Player和Monster两个字段。

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    golang--单元测试综合实例

    实例说明:(1)一个Monster结构体,字段Name,Age,Skill(2)Monster有一个Store方法,可以将一个Monster对象序列化后保存在文件中;(3)Monster有一个ReStore main.gopackage main import ( bufio encodingjson fmt io os) type monster struct { Name string `json:name ` Age int `json:age` Skill string `json:skill`} func (m *monster) store() byte(data), &m) if err ! fmt.Println(mon)}main_test.gopackage main import ( fmt testing) func TestStore(t *testing.T) { m := &monster } data := m.reStore(str) fmt.Println(data) t.Logf(反序列化成功)} func TestWriteFile(t *testing.T) { m := &monster

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    怪物补刀贪心算法

    The i-th monster has hi health points (hp). opponent go the second monster and fight it till his death, and so on. The fight with a monster happens in turns.轮流攻击怪兽。You hit the monster by a hp. Your opponent hits the monster by b hp. and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster

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    Laravel之容器(实战)背景设计客户端代码总结参考

    客户端代码在客户端使用,我们先给出没有使用容器的方式: public function test_use_sword() { 生成怪物 $monster1 = new Monster( 小怪A, 50 ); $monster2 = new Monster( 小怪B, 50 ); $monster3 = new Monster( 关主, 200 ); $monster4 = new Monster( 最终 new Container(); $container->bind(GameDemoIAttackStrategy,GameDemoWoodSword); 生成怪物 $monster1 = new Monster ( 小怪A, 50 ); $monster2 = new Monster( 小怪B, 50 ); $monster3 = new Monster( 关主, 200 ); $monster4 = new Monster( 最终Boss, 1000 ); 生成角色 ** * @var Role * $role = $container->make(GameDemoRole,); $container->rebinding

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    Java中的深复制和浅赋值

    浅复制下面的实例中,我们创建一个原始类Monster,调用对象的clone方法复制一个新的对象。 注意:要调用对象的clone方法,需要让类实现Cloneable接口,并重写clone方法public class Monster implements Cloneable{ private String name; 怪物名称 private int level; 怪物级别 private Point position; 怪物坐标 public Monster(String name, int level monster1 = new Monster(史莱克, 99); 通过clone方法,将对象monster1中的值复制到对象monster2 Monster monster2 = (Monster)monster1 monster1 = new Monster(史莱克, 99); 调用自定义的deepClone方法(深复制),将对象monster1中的值复制到对象monster2 Monster monster2

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    综合案例

    _mp class Monster(Fighter): 小怪兽 __slots__ = (_name, _hp) def attack(self, other): other.hp -= randint _hp def is_any_alive(monsters): 判断有没有小怪兽是活着的 for monster in monsters: if monster.alive > 0: return True = monsters if monster.alive > 0: return monster def display_info(ultraman, monsters): 显示奥特曼和小怪兽的信息 print (ultraman) for monster in monsters: print(monster, end=) def main(): u = Ultraman(骆昊, 1000, 120) m1 = Monster(舒小玲, 250) m2 = Monster(白元芳, 500) m3 = Monster(王大锤, 750) ms = fight_round = 1 while u.alive

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    Head First设计模式——原型模式和访问者模式

    原型模式中有三个角色:原型角色:定义用于复制现有实例来生成新实例的方法(Monster)。 实现代码:①原型角色1 public interface Monster2 {3 public Monster Clone();4 }②具体原型角色 1 public class WellKnowMonster : Monster 2 { 3 public Monster Clone() 4 { 5 Monster clone = JsonConvert.DeserializeObject(JsonConvert.SerializeObject Clone()14 {15 Monster clone = JsonConvert.DeserializeObject(JsonConvert.SerializeObject(this));16 return monster) { 5 monsterDic.Add(key, monster); 6 } 7 public Monster GetMonster(string key) { 8 Monster monster

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    Golang的垂直组合思维——type embedding

    struct { 9 Creature Monster继承自Monster10}这样做的好处就是,Monster直接可以调用到Creature里的方法,Creature直接可以调用Object里的方法 但是,Go中没有基类指针指向派生类对象,不可以Object指向一个Monster对象,调用Monster中的方法。 而我们实际上在很多地方需要这种抽象类型机制,比如存储需要存Creature类型,使用的时候再具体用Monster类型方法。 现在一个东西实现Object,如果它是Monster,那么一定是Creature。 最后我们给出Monster的代码,可以发现,他只需要实现自己独有的方法即可。

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    go语言chan 和 routine活用

    . * package monster monster 以小写命名,标识该类无法直接创建,这里由 monster.MonsterNew 创建 type monster struct { Id int Hp int } ** * Created by Administrator on 13-12-20. * package monster import ( sync fmt os time mathrand = make(map*monster, 100) 怪物map lock var monstersLock sync.Mutex = sync.Mutex{} 怪物血量命令通道 var monstersChan chan *MonsterHpChangeCommand, 100) 创建一个怪物,并且将他添加到“怪物map”并且为其注册“怪物命令通道map” func MonsterNew(id int) *monster { monstersLock.Lock() defer monstersLock.Unlock() m := &monster{id, 99999} monsters = m fmt.Fprintf(

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    Codeforces Round #547 (Div. 3)E. Superhero Battle

    secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputA superhero fights with a monster It means that before the battle the monster has HH hit points. Print the first minute after which the monster dies. The monster dies if its hp is less than or equal to 00. Otherwise, print the positive integer kk such that kk is the first minute after which the monster is

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    (三十九)golang--反序列化

    import ( encodingjson fmt) type monster struct { Name string `json:name` Age int `json:age` Birthday skill`} func unmarshalStruct() { str := {name:牛魔王,age:10,birthday:1994-09-18,sal:10000,skill:牛魔拳} var monster monster err := json.Unmarshal(interface{} err := json.Unmarshal(}, + {age:30,name:猪八戒}] var a interface = nil { fmt.Println(unmarshal error=, err) } fmt.Printf(反序列化后 monster=%vn, a)} func main() { unmarshalStruct

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    程序员进阶之算法练习(十一)有感而发

    hdu 5811题目链接 题目大意: 有n只monster; 输入一个矩阵mat,表示monster之间的关系; mat=1表示i怪物能打败j怪物; mat=0表示i怪物打不过j怪物; 注意,mat=0 输入m,再输入m个数字,表示挑选出m只monster组成T1队,剩下n-m只monster组成T2队,询问T1队与T2队是否合法 (合法是指存在某种排列,排列中任意一只monster都能打败他右边所有monster 如果合法,最多从T2队中能选出几只monster插入到T1中,并保证T1合法? 代码实现 题目解析: 要求1,可以用拓扑排序;(O(n^2)) 要求2,先按照T2队从大到小的顺序,求出T2队中每个monster对应在T1队的位置pos,这样就转换成求pos数组的最长不下降子序列;( O(n^2)的dp) 注意1、2、2、3这种序列是合法的,因为原来T2本来就满足顺序要求(排列中任意一只monster都能打败他右边所有monster); hdu 5815题目链接 题目大意: 有一个n

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    Golang之旅35-struct4

    B a = A(b) 转换:结构体的字段要完全相同 a = b 报错 fmt.Println(a, b)}package mainimport ( fmt encodingjson) type Monster 创建一个Monster 变量 monster := Monster{牛魔王, 500, 芭蕉扇~} 2. 将monster变量序列化为一个json格式字符串 json.Marshal 函数中使用到了反射机制 jsonstr, err := json.Marshal(monster) if err !

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    2019 ICPC 银川网络赛 H. Fight Against Monsters

    The health point of the ii-th monster was HP_iHPi​, and its attack value was ATK_iATKi​.They fought in Then he selected a monster and attacked it. The monster would suffer the damage of kk (its health point would decrease by kk) which was the times the damage of Huriyyahs second attack to this monster was 22, the third time to this monster was 33, If at some time, the health point of a monster was less than or equal to zero, it died.

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    (四十五)golang--反射

    package main import ( fmt reflect) type monster struct { Name string `json:name` Age int `json:monster_age ` Score float32 Sex string} func (m monster) Print() { fmt.Println(----strat----) fmt.Println() fmt.Println (----end----)} func (m monster) GetSum(n1, n2 int) int { return n1 + n2} func (m monster) Set(name string 所以这里第二个方法是Print rval.Elem().Method(1).Call(nil) 定义一个reflect.Value切片 var params .Int()) } func main() { var a monster = monster{ Name: tom, Age: 400, Score: 20.9, } testStruct(&a) fmt.Println(a)}要在反射中修改结构体的字段的值,必须传入的是地址

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    SpringBoot-02-之参数传递

    properties.MonsterProperties@Component org.springframework.stereotype.Component;@ConfigurationProperties(prefix = monster 获取配置文件组成员 @RestController public class HelloSpringBoot { @Autowired 自动创建对象 private MonsterProperties monster ; @GetMapping(monster) public String say() { return monster.toString(); } }访问:http:localhost:8080monster

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    Python随笔(四)抽象语法树AST 原

    next_instrn_nchildren = 0; n->n_child = NULL; return n;}下面给出Python自带的AST例子,去观察构建出来的树import astMonster =class Monster self.level=0 self.hp=1000 self.boom= def eat(self,frut): self.hp+=1 def howl(self): print(Ao uuuuuuuuuuuuuu)monster =Monster()monster.howl()if __name__==__main__: # cm = compile(Monster, , exec) # exec (cm) r_node = ast.parse (Monster) print(ast.dump(r_node))通过compile我们可以编译Python字符串执行字串的内容 ?

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    图数据库查询语言

    pluto, age, 4000, type, god)Vertex nemean = graph.addVertex(T.label, character, name, nemean, type, monster )Vertex hydra = graph.addVertex(T.label, character, name, hydra, type, monster)Vertex cerberus = graph.addVertex (T.label, character, name, cerberus, type, monster)Vertex tartarus = graph.addVertex(T.label, location

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