zoukankan      html  css  js  c++  java
  • codeforces487A

    Fight the Monster

     CodeForces - 487A 

    A monster is attacking the Cyberland!

    Master Yang, a braver, is going to beat the monster. Yang and the monster each have 3 attributes: hitpoints (HP), offensive power (ATK) and defensive power (DEF).

    During the battle, every second the monster's HP decrease by max(0, ATKY - DEFM), while Yang's HP decreases by max(0, ATKM - DEFY), where index Y denotes Master Yang and index M denotes monster. Both decreases happen simultaneously Once monster's HP ≤ 0 and the same time Master Yang's HP > 0, Master Yang wins.

    Master Yang can buy attributes from the magic shop of Cyberland: h bitcoins per HPa bitcoins per ATK, and d bitcoins per DEF.

    Now Master Yang wants to know the minimum number of bitcoins he can spend in order to win.

    Input

    The first line contains three integers HPY, ATKY, DEFY, separated by a space, denoting the initial HPATK and DEF of Master Yang.

    The second line contains three integers HPM, ATKM, DEFM, separated by a space, denoting the HPATK and DEF of the monster.

    The third line contains three integers h, a, d, separated by a space, denoting the price of HPATK and DEF.

    All numbers in input are integer and lie between 1 and 100 inclusively.

    Output

    The only output line should contain an integer, denoting the minimum bitcoins Master Yang should spend in order to win.

    Examples

    Input
    1 2 1
    1 100 1
    1 100 100
    Output
    99
    Input
    100 100 100
    1 1 1
    1 1 1
    Output
    0

    Note

    For the first sample, prices for ATK and DEF are extremely high. Master Yang can buy 99 HP, then he can beat the monster with 1 HP left.

    For the second sample, Master Yang is strong enough to beat the monster, so he doesn't need to buy anything.

    sol:一开始想了好几个SB贪心都被自己否定掉了,猛然发现数据范围小的不像话,暴力枚举n2即可

    #include <bits/stdc++.h>
    using namespace std;
    typedef int ll;
    inline ll read()
    {
        ll s=0;
        bool f=0;
        char ch=' ';
        while(!isdigit(ch))
        {
            f|=(ch=='-'); ch=getchar();
        }
        while(isdigit(ch))
        {
            s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
        }
        return (f)?(-s):(s);
    }
    #define R(x) x=read()
    inline void write(ll x)
    {
        if(x<0)
        {
            putchar('-'); x=-x;
        }
        if(x<10)
        {
            putchar(x+'0'); return;
        }
        write(x/10);
        putchar((x%10)+'0');
        return;
    }
    #define W(x) write(x),putchar(' ')
    #define Wl(x) write(x),putchar('
    ')
    const int inf=0x3f3f3f3f;
    struct Record
    {
        int Hp,Atk,Def;
    }a,b,Buy;
    int main()
    {
        int i,j,k,ans=inf;
        R(a.Hp); R(a.Atk); R(a.Def);
        R(b.Hp); R(b.Atk); R(b.Def);
        R(Buy.Hp); R(Buy.Atk); R(Buy.Def);
        for(i=1;i<=b.Hp;i++)//几回合干掉怪兽
        {
            int ATK=(b.Hp/i+(bool)(b.Hp%i))+b.Def;
            for(j=0;j<=b.Atk;j++)//增加几点防御
            {
                int DEF=a.Def+j;
                int Sum=max(ATK-a.Atk,0)*Buy.Atk+j*Buy.Def;
                int Hp=max(b.Atk-DEF,0)*i+1;
                ans=min(ans,Sum+max(Hp-a.Hp,0)*Buy.Hp);
            }
        }
        Wl(ans);
        return 0;
    }
    /*
    Input
    1 2 1
    1 100 1
    1 100 100
    Output
    99
    
    Input
    100 100 100
    1 1 1
    1 1 1
    Output
    0
    */
    View Code
  • 相关阅读:
    Epplus导出excel
    访问GitHub需要修改hosts
    如何将你的.Net Core程序部署成为服务
    生成雪花Id类
    文件操作帮助类
    工作流-WikeFlow
    《C语言进阶剖析》课程目录
    《C++深度解析》课程目录
    USB URB的status及其代表的意义
    数据结构优秀博文整理
  • 原文地址:https://www.cnblogs.com/gaojunonly1/p/10745685.html
Copyright © 2011-2022 走看看