【杭电2015年12月校赛I】【模拟水题】The Magic Tower 战士打魔王 能否打死它

The Magic Tower

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2434    Accepted Submission(s): 630

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

W100 1000 900100 1000 900B100 1000 900100 1000 900

 

Sample Output

Warrior winsWarrior loses

 

#include<stdio.h>#include<algorithm>#include<ctype.h>#include<string.h>using namespace std;int casenum,casei;typedef long long LL;const int N=105;const int M=105*105;int n,m;struct A{    int x,y,z;    bool operator < (const A& b)const    {        return z>b.z;    }}a[M];int f[N];int find(int x){    if(f[x]==x)return x;    f[x]=find(f[x]);    return f[x];}char C;int hp[2];int att[2];int def[2];int hurt[2];bool solve(){    hurt[0]=att[0]-def[1];    hurt[1]=att[1]-def[0];    if(hurt[0]<=0)return 0;    if(hurt[1]<=0)return 1;    int p=(C=='W'?0:1);    while(1)    {        hp[p^1]-=hurt[p];        if(hp[p^1]<=0)        {            return p==0;        }        p^=1;    }}int main(){    while(~scanf(" %c",&C))    {        for(int i=0;i<2;++i)scanf("%d%d%d",&hp[i],&att[i],&def[i]);        puts(solve()?"Warrior wins":"Warrior loses");    }    return 0;}/*【题意】又到了战士和魔王决斗的时刻了。两人轮流攻击对方，生命值<=0为死亡。告诉你谁先手，告诉你战士和魔王的hp,attack,defence，伤害=攻击者的attack-防御者的defence（数值都是[1,10000]之间的整数）让你输出战士是否能最终击杀魔王可以输出Warrior wins否则输出Warrior loses【类型】模拟 水题【分析】看代码就好了，那里逻辑很清楚。注意，使用结构有序的写法，写起来也会超级方便哦。【时间复杂度&&优化】最坏情况下的时间复杂度为O（hp[0]+hp[1]）*/