Problem Description

Like most of the RPG (role play game), “The Magic Tower” is a game about how a warrior saves the princess. 
After killing lots of monsters, the warrior has climbed up the top of the magic tower. There is a boss in front of him. The warrior must kill the boss to save the princess.
Now, the warrior wants you to tell him if he can save the princess.
Input
There are several test cases.
For each case, the first line is a character, “W” or “B”, indicating that who begins to attack first, ”W” for warrior and ”B” for boss. They attack each other in turn. 
The second line contains three integers, W_HP, W_ATK and W_DEF. (1<=W_HP<=10000, 0<=W_ATK, W_DEF<=65535), indicating warrior’s life point, attack value and defense value. 
The third line contains three integers, B_HP, B_ATK and B_DEF. (1<=B_HP<=10000, 0<=B_ATK, B_DEF<=65535), indicating boss’s life point, attack value and defense value. 
Note: warrior can make a damage of (W_ATK-B_DEF) to boss if (W_ATK-B_DEF) bigger than zero, otherwise no damage. Also, boss can make a damage of (B_ATK-W_DEF) to warrior if (B_ATK-W_DEF) bigger than zero, otherwise no damage. 
Output
For each case, if boss’s HP first turns to be smaller or equal than zero, please print ”Warrior wins”. Otherwise, please print “Warrior loses”. If warrior cannot kill the boss forever, please also print ”Warrior loses”.
Sample Input
W
100 1000 900
100 1000 900
B
100 1000 900
100 1000 900
Sample Output
Warrior wins
Warrior loses
 模拟题
理解题意就好
这里我先判断可以胜利和失败的情况‘
再讨论谁先攻击的情况
#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
int n;
int main ()
{
string s;
while(cin>>s)
{
int W_HP,W_ATK,W_DE;
int B_HP,B_ATK,B_DE; int F;
int D;
cin>>W_HP>>W_ATK>>W_DE;
cin>>B_HP>>B_ATK>>B_DE;
int CA=W_ATK-B_DE;
int CB=B_ATK-W_DE;
if(W_ATK<=B_DE)
{
puts("Warrior loses");
}
else if(B_ATK<=W_DE&&W_ATK>B_DE)
{
puts("Warrior wins");
}
else if(W_ATK>B_DE&&B_ATK>W_DE)
{
if(s[0]=='W')
{
int S_1=B_HP;
int S_2=W_HP;
while(S_1>0&&S_2>0)
{
S_1-=CA;
S_2-=CB;
}
if(S_1<=0)
{
puts("Warrior wins");
}
else if(S_2<=0&&S_1>0)
{
puts("Warrior loses");
}
}
else
{
int S_1=B_HP;
int S_2=W_HP;
while(S_1>0&&S_2>0)
{
S_2-=CB;
S_1-=CA;
}
if(S_2<=0)
{
puts("Warrior loses");
}
else if(S_2>0&&S_1<=0)
{
puts("Warrior wins");
}
}
}
}
return 0;
}

  

HDU计算机学院大学生程序设计竞赛(2015’12)The Magic Tower的更多相关文章

  1. hdu 计算机学院大学生程序设计竞赛(2015’11)

    搬砖 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submissi ...

  2. HDU计算机学院大学生程序设计竞赛(2015’12)Happy Value

    Problem Description In an apartment, there are N residents. The Internet Service Provider (ISP) want ...

  3. HDU计算机学院大学生程序设计竞赛(2015’12)The Country List

    Problem Description As the 2010 World Expo hosted by Shanghai is coming, CC is very honorable to be ...

  4. 计算机学院大学生程序设计竞赛(2015’11)1005 ACM组队安排

    1005 ACM组队安排 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Pro ...

  5. 计算机学院大学生程序设计竞赛(2015’12)Study Words

    Study Words Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Tota ...

  6. 计算机学院大学生程序设计竞赛(2015’12)Polygon

    Polygon Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Su ...

  7. 计算机学院大学生程序设计竞赛(2015’12)The Country List

    The Country List Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  8. 计算机学院大学生程序设计竞赛(2015’12) 1008 Study Words

    #include<cstdio> #include<cstring> #include<map> #include<string> #include&l ...

  9. 计算机学院大学生程序设计竞赛(2015’12) 1009 The Magic Tower

    #include<cmath> #include<cstdio> #include<cstring> #include<algorithm> using ...

随机推荐

  1. 重命名File

    File completeFile = new File(mFilePath + mFileName); if (completeFile.exists()) { File fileWithSuffi ...

  2. [poj2449]Remmarguts' Date(K短路模板题,A*算法)

    解题关键:k短路模板题,A*算法解决. #include<cstdio> #include<cstring> #include<algorithm> #includ ...

  3. div+css学习笔记一(转)

    div+css学习笔记一 (2011-05-12 07:32:08) 标签: div css 居中 背景图片 ie6 ie7 margin 杂谈 分类: 网页制作 1.IE6中用了float之后导致m ...

  4. Codeforces 464E The Classic Problem (最短路 + 主席树 + hash)

    题意及思路 这个题加深了我对主席树的理解,是个好题.每次更新某个点的距离时,是以之前对这个点的插入操作形成的线段树为基础,在O(logn)的时间中造出了一颗新的线段树,相比直接创建n颗线段树更省时间. ...

  5. ???SpringMVC_03 利用SpringMVC提供的过滤器解决浏览器请求参数的乱码问题

    1 响应乱码问题 在启用mvc注解的配置中添加一个转换器配置 <?xml version="1.0" encoding="UTF-8"?> < ...

  6. Spring_02 注入类型值、利用引用注入类型值、spring表达式、与类相关的注解、与依赖注入相关的注解、注解扫描

    注意:注入基本类型值在本质上就是依赖注入,而且是利用的set方式进行的依赖注入 1 注入基本类型的值 <property name="基本类型的成员变量名" value=&q ...

  7. php学习笔记-for循环

    for(init;condition;statement) { func(); } for循环的执行逻辑是先执行一次init语句,然后判断condition是否为true,是则执行func(),再执行 ...

  8. Linux 编译安装内核

    一.简介 内核,是一个操作系统的核心.它负责管理系统的进程.内存.设备驱动程序.文件和网络系统,决定着系统的性能和稳定性.Linux作为一个自由软件,在广大爱好者的支持下,内核版本不断更新.新的内核修 ...

  9. 在centos上安装sequoaidb的php驱动

    1:搭建PHP的运行环境 yum install httpd httpd-devel yum install php   php-devel yum install php-gd php-imap p ...

  10. Java之命令模式(Command Pattern)

    转自:http://www.cnblogs.com/devinzhang/archive/2012/01/06/2315235.html 1.概念 将来自客户端的请求传入一个对象,从而使你可用不同的请 ...