• Educational Codeforces Round 34 B. The Modcrab【模拟/STL】


    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 bya1) 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.

    【分析】:STL还是用的太少了···C++还要再学一波

    【代码】:

    #include <bits/stdc++.h>
    using namespace std;
    #define ll long long
    #define oo 10000000
    const int mod = 1e6;
    
    int main()
    {
        vector<string> v;
        int a1,b1,h1;
        int a2,b2;
        int j=0;
        int f[500];
        scanf("%d%d%d",&a1,&b1,&h1);
        scanf("%d%d",&a2,&b2);
        while(1)
        {
            if(b1>=a2) //一击必杀就跳出···因为怪兽你已经死了
            {
                v.push_back("STRIKE");
                break;
            }
            else
            {
                if(a1<=b2)
                {
                    a1 += h1; //补血
                    a1 -= b2; //补完还是要被打 怪兽不会等你补就不打了
                    v.push_back("HEAL");
                }
                else
                {
                    a1 -= b2; //我的血少了
                    a2 -= b1; //怪兽的血少了
                    v.push_back("STRIKE");
                }
            }
        }
        printf("%d
    ",(int)v.size());
        
        for(auto &i:v)
        {
            printf("%s
    ",i.c_str());
        }
        return 0;
    }
    View Code
  • 相关阅读:
    8086 CPU 寄存器
    python中 * 的用法
    字典的相应操作
    tesseract学习记录
    C学习之路2012.8.28
    函数库管理
    2013.3.19C++浏览记录。。。
    自动生成makefile文件学习
    整理做过的东西(电子警察)
    基于zed的tesseract移植过程记录
  • 原文地址:https://www.cnblogs.com/Roni-i/p/8030628.html
Copyright © 2020-2023  润新知