计算机学院大学生程序设计竞赛(2015’12)The Magic Tower(水~~~)
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The Magic Tower
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2398 Accepted Submission(s): 618
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.
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.
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 <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <queue>#include <stack>#include <set>#include <map>#include <list>#define LL long long#define INF 0x3f3f3f3fusing namespace std;int main(){ char ch; int a1,b1,c1; int a2,b2,c2; while(cin>>ch) { cin>>a1>>b1>>c1; cin>>a2>>b2>>c2; if(c2 - b1 >= 0) cout<<"Warrior loses"<<endl; else { if(c1 - b2 >= 0) cout<<"Warrior wins"<<endl; else { if(ch == 'W') { while(1) { a2 -= b1; if(a2 <= 0) { cout<<"Warrior wins"<<endl; break; } a1 -= b2; if(a1 <= 0) { cout<<"Warrior loses"<<endl; break; } } } else { while(1) { a1 -= b2; if(a1 <= 0) { cout<<"Warrior loses"<<endl; break; } a2 -= b1; if(a2 <= 0) { cout<<"Warrior wins"<<endl; break; } } } } } } return 0;}
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