The Modcrab_Codeforces

原题链接:http://codeforces.com/contest/903/problem/B

B. The Modcrab
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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 c1Vova'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 h1a1c1 (1 ≤ h1, a1 ≤ 1002 ≤ c1 ≤ 100) — Vova's health, Vova's attack power and the healing power of a potion.

The second line contains two integers h2a2 (1 ≤ h2 ≤ 1001 ≤ 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.


重点:每次Vova攻击之前先判断是否在本轮结束之后Vova的血量会降到0及其以下,如果是,则进行HEAL操作,否则进行STRIKE操作(注意:如果Vova攻击之后Modcrab已死,Vova则不会再收到攻击,不需要进行HEAL操作)


#include<cstdio>

int main()
{
    int vhealth, vattack, portion;
    scanf("%d%d%d", &vhealth, &vattack, &portion);
    int mhealth, mattack;
    scanf("%d%d", &mhealth, &mattack);
    int cnt = 0;
    int s[10000] = {0};
    while(mhealth > 0)
    {
        if(vhealth <= mattack && vattack < mhealth)
        {
            s[cnt] = 1;
            vhealth += portion;
            vhealth -= mattack;
        }
        else
        {
            s[cnt] = 2;
            mhealth -= vattack;
            vhealth -= mattack;
        }
        cnt++;
    }
    printf("%d\n", cnt);
    for(int i = 0; i < cnt; i++)
    {
        if(s[i] == 1)
            printf("HEAL\n");
        else if(s[i] == 2)
            printf("STRIKE\n");
    }
    return 0;
}

关于Codeforces上的问题'Trail',目前提供的参考资料中并未直接提及该问题的具体解法或讨论[^1]。然而,在处理类似平台上的编程挑战时,通常会遵循特定的方法论来解决问题。 对于未具体描述的问题'Trail',假设这是一个涉及路径遍历或是图结构中的轨迹计算等问题,一般解决方案可能涉及到深度优先搜索(DFS)、广度优先搜索(BFS)或者是动态规划等技术。这些方法能够有效地探索所有可能性并找到最优解。 考虑到Codeforces平台上许多问题的特点,解决这类题目往往还需要注意边界条件以及输入数据范围的影响。编写代码前应仔细阅读题目说明,确保理解所有的约束条件和特殊案例。 下面是一个简单的Python实现例子,用于展示如何通过深度优先搜索算法在一个假定的网格环境中寻找从起点到终点的有效路径: ```python def dfs(grid, start, end): rows, cols = len(grid), len(grid[0]) visited = set() def explore(r, c): if (r < 0 or r >= rows or c < 0 or c >= cols or grid[r][c] == '#' or (r,c) in visited): return False if (r, c) == end: return True visited.add((r, c)) directions = [(0, 1), (1, 0), (-1, 0), (0, -1)] for dr, dc in directions: next_r, next_c = r + dr, c + dc if explore(next_r, next_c): return True return False return explore(*start) # Example usage with a simple maze represented as a list of strings. maze = [ '..#.##', '#...#.', '#####.' ] print(dfs(maze, (0, 0), (2, 5))) # Output should be True based on this example layout. ``` 此段代码展示了利用递归方式执行深度优先搜索的过程,适用于某些类型的‘Trail’类问题。当然实际应用中还需根据具体的题目要求调整逻辑细节。
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