cf 903B - The Modcrab

最新推荐文章于 2020-12-19 19:35:54 发布

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本文介绍了一个游戏战斗模拟问题，玩家角色通过攻击和适时使用回血药剂战胜怪物。文章详细阐述了模拟过程，并提供了一段C++代码实现，用于计算最小胜利回合数。

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Vova is again playing some computer game, now an RPG. In the game Vova's character received a quest: to slay the fearsome monster called Modcrab.

After two hours of playing the game Vova has tracked the monster and analyzed its tactics. The Modcrab has h2 health points and an attack power of a2. Knowing that, Vova has decided to buy a lot of strong healing potions and to prepare for battle.

Vova's character has h1 health points and an attack power of a1. Also he has a large supply of healing potions, each of which increases his current amount of health points by c1 when Vova drinks a potion. All potions are identical to each other. It is guaranteed that c1 > a2.

The battle consists of multiple phases. In the beginning of each phase, Vova can either attack the monster (thus reducing its health by a1) or drink a healing potion (it increases Vova's health by c1; Vova's health can exceed h1). Then, if the battle is not over yet, the Modcrab attacks Vova, reducing his health by a2. The battle ends when Vova's (or Modcrab's) health drops to 0 or lower. It is possible that the battle ends in a middle of a phase after Vova's attack.

Of course, Vova wants to win the fight. But also he wants to do it as fast as possible. So he wants to make up a strategy that will allow him to win the fight after the minimum possible number of phases.

Help Vova to make up a strategy! You may assume that Vova never runs out of healing potions, and that he can always win.

Input

The first line contains three integers h1, a1, c1 (1 ≤ h1, a1 ≤ 100, 2 ≤ c1 ≤ 100) — Vova's health, Vova's attack power and the healing power of a potion.

The second line contains two integers h2, a2 (1 ≤ h2 ≤ 100, 1 ≤ a2 < c1) — the Modcrab's health and his attack power.

Output

In the first line print one integer n denoting the minimum number of phases required to win the battle.

Then print n lines. i-th line must be equal to HEAL if Vova drinks a potion in i-th phase, or STRIKE if he attacks the Modcrab.

The strategy must be valid: Vova's character must not be defeated before slaying the Modcrab, and the monster's health must be 0 or lower after Vova's last action.

If there are multiple optimal solutions, print any of them.

Examples

input
10 6 100
17 5

output
4
STRIKE
HEAL
STRIKE
STRIKE

input
11 6 100
12 5

output
2
STRIKE
STRIKE

Note

In the first example Vova's character must heal before or after his first attack. Otherwise his health will drop to zero in 2 phases while he needs 3 strikes to win.

In the second example no healing needed, two strikes are enough to get monster to zero health and win with 6 health left.

题意:X,Y两人 PK. X攻击为a1,血量为h1,可吃药回血c(无限次).

Y攻击为a2(保证小于c == X必胜),血量为h2,不可回血

现在要求出最少的回合数,使得X赢Y

思路:模拟过程,陷阱在于我们考虑X能不能承受B的下一次攻击的时候,要先提前判断,X当前一次攻击a1>h2?

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define bug1 cout <<"bug1"<<endl
#define bug2 cout <<"bug2"<<endl
#define bug3 cout <<"bug3"<<endl
#define long long ll;
using namespace std;

vector <string> ans;

int main(void){
    int h1,a1,c1,h2,a2;
    cin >>h1>>a1>>c1>>h2>>a2;
    while(h1>0 && h2>0){
        if(a1>=h2){
//            cout <<"1"<<endl;
            ans.push_back("STRIKE");
            break;
        }
        if(h1>a2){
            ans.push_back("STRIKE");
            h1-=a2;
            h2-=a1;
        }
        else{
            ans.push_back("HEAL");
            h1+=c1;
            h1-=a2;
        }
//        cout <<"h1=" <<h1<<" h2="<<h2<< " flag=" << flag<<endl;;
    }
    cout << ans.size() <<endl;
    for(register int i=0;i<ans.size();i++)
        cout <<ans[i]<<endl;
}