Educational Codeforces Round 34 B. The Modcrab【模拟/STL】
1 second
256 megabytes
standard input
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.
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.
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.
10 6 100
17 5
4
STRIKE
HEAL
STRIKE
STRIKE
11 6 100
12 5
2
STRIKE
STRIKE
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=;
int f[];
scanf("%d%d%d",&a1,&b1,&h1);
scanf("%d%d",&a2,&b2);
while()
{
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\n",(int)v.size()); for(auto &i:v)
{
printf("%s\n",i.c_str());
}
return ;
}
Educational Codeforces Round 34 B. The Modcrab【模拟/STL】的更多相关文章
- Educational Codeforces Round 34 (Rated for Div. 2) A B C D
Educational Codeforces Round 34 (Rated for Div. 2) A Hungry Student Problem 题目链接: http://codeforces. ...
- Educational Codeforces Round 34 (Rated for Div. 2) B题【打怪模拟】
B. The Modcrab Vova is again playing some computer game, now an RPG. In the game Vova's character re ...
- Educational Codeforces Round 34 D. Almost Difference【模拟/stl-map/ long double】
D. Almost Difference time limit per test 2 seconds memory limit per test 256 megabytes input standar ...
- Educational Codeforces Round 34 C. Boxes Packing【模拟/STL-map/俄罗斯套娃】
C. Boxes Packing time limit per test 1 second memory limit per test 256 megabytes input standard inp ...
- Educational Codeforces Round 2 A. Extract Numbers 模拟题
A. Extract Numbers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/600/pr ...
- Educational Codeforces Round 11B. Seating On Bus 模拟
地址:http://codeforces.com/contest/660/problem/B 题目: B. Seating On Bus time limit per test 1 second me ...
- Educational Codeforces Round 34 (Rated for Div. 2)
A. Hungry Student Problem time limit per test 1 second memory limit per test 256 megabytes input sta ...
- Educational Codeforces Round 34
F - Clear The Matrix 分析 题目问将所有星变成点的花费,限制了行数(只有4行),就可以往状压DP上去靠了. \(dp[i][j]\) 表示到第 \(i\) 列时状态为 \(j\) ...
- Educational Codeforces Round 34 (Rated for Div. 2) D - Almost Difference(高精度)
D. Almost Difference Let's denote a function You are given an array a consisting of n integers. You ...
随机推荐
- ZOJ Monthly, January 2018 训练部分解题报告
A是水题,此处略去题解 B - PreSuffix ZOJ - 3995 (fail树+LCA) 给定多个字符串,每次询问查询两个字符串的一个后缀,该后缀必须是所有字符串中某个字符串的前缀,问该后缀最 ...
- TCP/IP网络编程之套接字类型与协议设置
套接字与协议 如果相隔很远的两人要进行通话,必须先决定对话方式.如果一方使用电话,另一方也必须使用电话,而不是书信.可以说,电话就是两人对话的协议.协议是对话中使用的通信规则,扩展到计算机领域可整理为 ...
- SetUnhandledExceptionFilter
SetUnhandledExceptionFilter 设置未捕获异常处理 通常windows程序长时间运行,会发生各种问题,例如访问异常,内存溢出,堆栈破坏等. 这时候通常希望程序自己能增加处理,而 ...
- Python框架之Django学习笔记(十)
又是一周周末,如约学习Django框架.在上一次,介绍了MVC开发模式以及Django自己的MVT开发模式,此次,就从数据处理层Model谈起. 数据库配置 首先,我们需要做些初始配置:我们需要告诉D ...
- 【Set Matrix Zeros】cpp
题目: Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place. cl ...
- 详细的Windows下安装 binwalk
1. https://github.com/ReFirmLabs/binwalk到这里下载binwalk,下载后解压. 2. 找到下载后的文件夹, 在这里要进行安装步骤,一边按着shift,一边按着鼠 ...
- 502 Bad Gateway 怎么解决?
出现502的原因是:对用户访问请求的响应超时造成的 服务端解决办法: 1.提高 Web 服务器的响应速度,也即减少内部的调用关系,可以把需要的页面.素材或数据,缓存在内存中,可以是专门的缓存服务器 , ...
- 踩坑 PHP Fatal Error Failed opening required File
使用 require 引用文件时,报错如下: require 'https://dev.ryan.com/test.php'; [Sat Mar 19 23:10:50 2011] [warn] mo ...
- scp -v 查看具体的过程
前几天跟同事讨论scp 多个文件和 scp多个文件夹的压缩包那个快. 老大说,压缩包快,压缩包传输可以避免每个文件的重建连接,不过文件系统的遍历.目录创建.检验会有一些开销. 他建议我scp -v看下 ...
- Android数据储存之SQLiteDatabase SQLiteOpenHelper类的简单使用
SQLiteOpenHelper 简介: SQLiteOpenHelper是一个借口!所以不能直接实例化!那我们想要得到SQLiteOpenHelper对象就需要实现该接口!创建该接口的实现类对象! ...