zoukankan      html  css  js  c++  java
  • 4321(贪心)

    Problem Description

    When playing DotA with god-like rivals and pig-like team members, you have to face an embarrassing situation: All your teammates are killed, and you have to fight 1vN.

    There are two key attributes for the heroes in the game, health point (HP) and damage per shot (DPS). Your hero has almost infinite HP, but only 1 DPS.

    To simplify the problem, we assume the game is turn-based, but not real-time. In each round, you can choose one enemy hero to attack, and his HP will decrease by 1. While at the same time, all the lived enemy heroes will attack you, and your HP will decrease by the sum of their DPS. If one hero's HP fall equal to (or below) zero, he will die after this round, and cannot attack you in the following rounds.

    Although your hero is undefeated, you want to choose best strategy to kill all the enemy heroes with minimum HP loss.

    Input

    The first line of each test case contains the number of enemy heroes N (1 <= N <= 20). Then N lines followed, each contains two integers DPSi and HPi, which are the DPS and HP for each hero. (1 <= DPSi, HPi <= 1000)

    Output

    Output one line for each test, indicates the minimum HP loss.

    Sample Input

    1
    10 2
    2
    100 1
    1 100
    

    Sample Output

    20
    201
    

    #include<stdio.h>
    struct n
    {
        int hp,dps;
    };
    int main()
    {
        int n,i,j;
        __int64 sdps,sum;
         struct n  a[25],tem;
        while(scanf("%d",&n)==1)
        {
            sdps=0; sum=0;
            for(i=0;i<n;i++)
            {
                scanf("%d%d",&a[i].hp,&a[i].dps);
                sdps+=a[i].dps;
            }
    
            for(i=0;i<n;i++)
            for(j=i+1;j<n;j++)
            {
                if(a[i].dps*1.0/a[i].hp<a[j].dps*1.0/a[j].hp)
                {
                    tem=a[i];a[i]=a[j];a[j]=tem;
                }
            }
            for(i=0;i<n;i++)
            {
                sdps-=a[i].dps;
                sum+=a[i].hp*(sdps+a[i].dps);
            }
            printf("%I64d\n",sum);
        }
    }
    
    /*2
    30 1000
    1 50*/


     

  • 相关阅读:
    向MyEclipse添加Oracle数据库
    如何让搜索引擎抓取AJAX内容?
    XCode常用快捷键
    VMware Workstation 9上安装Mac OS X 10.8
    IOS学习第一篇——利用Xcode中的Interface Builder创建Hello World示例
    FM 101.7
    SqlServer游标操作
    添加COOKIE
    c#活动目录操作
    WCF服务调用方式
  • 原文地址:https://www.cnblogs.com/javawebsoa/p/3112988.html
Copyright © 2011-2022 走看看