The Modcrab (简单模拟)

The Modcrab

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.

code:

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main(){
    int h1,a1,c1;
    int h2,a2;
    cin >> h1 >> a1 >> c1;
    cin >> h2 >> a2;
    int attack = h2 / a1;
    int heal = 0;
    if(attack * a1 < h2) attack++;
    int be_attacked = attack - 1;
    int need = be_attacked * a2 - h1;
    if(need >= 0){
        int add = c1 - a2;
        heal = need / add;
        if(heal * add <= need) heal++;
    }
    printf("%d\n",heal+attack);
    while(heal--){
        printf("HEAL\n");
    }
    while(attack--){
        printf("STRIKE\n");
    }
    return 0;
}


MATLAB代码实现了一个基于多种智能优化算法优化RBF神经网络的回归预测模型,其核心是通过智能优化算法自动寻找最优的RBF扩展参数(spread),以提升预测精度。 1.主要功能 多算法优化RBF网络:使用多种智能优化算法优化RBF神经网络的核心参数spread。 回归预测:对输入特征进行回归预测,适用于连续值输出问题。 性能对比:对比不同优化算法在训练集和测试集上的预测性能,绘制适应度曲线、预测对比图、误差指标柱状图等。 2.算法步骤 数据准备:导入数据,随机打乱,划分训练集和测试集(默认7:3)。 数据归一化:使用mapminmax将输入和输出归一化到[0,1]区间。 标准RBF建模:使用固定spread=100建立基准RBF模型。 智能优化循环: 调用优化算法(从指定文件夹中读取算法文件)优化spread参数。 使用优化后的spread重新训练RBF网络。 评估预测结果,保存性能指标。 结果可视化: 绘制适应度曲线、训练集/测试集预测对比图。 绘制误差指标(MAE、RMSE、MAPE、MBE)柱状图。 十种智能优化算法分别是: GWO:灰狼算法 HBA:蜜獾算法 IAO:改进天鹰优化算法,改进①:Tent混沌映射种群初始化,改进②:自适应权重 MFO:飞蛾扑火算法 MPA:海洋捕食者算法 NGO:北方苍鹰算法 OOA:鱼鹰优化算法 RTH:红尾鹰算法 WOA:鲸鱼算法 ZOA:斑马算法
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值