Gym 100269E-Energy Tycoon

最新推荐文章于 2022-07-31 12:43:30 发布

原创最新推荐文章于 2022-07-31 12:43:30 发布 · 515 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Gym100269

贪心专栏收录该内容

2 篇文章

订阅专栏

本文介绍了一个计算机游戏“能源大亨”的策略模拟算法。玩家通过放置不同规模的发电厂来获得分数，目标是在有限的空间内实现最高得分。文章提供了一段C++代码实现，详细解释了如何在每回合中做出最优决策。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Input file: energy.in
Output file: energy.out
Time limit: 2 seconds
Memory limit: 256 megabytes
Little Vasya is playing a new computer game — turn-based strategy “Energy Tycoon”.
The rules of the game are quite simple:
• The board contains n slots arranged in a line.
• There are power plants, one power plant occupies one or two consecutive slots, and produces one
unit of energy.
• Each turn the game allows you to build one new power plant, you can put it on the board if you
wish. If there is no place for the new power plant, you can remove some older power plants.
• After each turn, the computer counts the amount of energy produced by the power plants on the
board and adds it to the total score.
这里写图片描述
Vasya already knows the types of power plant he will be able to build each turn. Now he wants to know,
what the maximum possible score he can get is. Can you help him?
Input
The first line of the input contains one integer n (1 ≤ n ≤ 100 000) — the number of slots on the board.
The second line contains the string s. The i-th character of the string is 1 if you can build one-slot power
plant at the i-th turn and the character is 2 if you can build two-slot power plant at the i-th turn. The
number of turns does not exceed 100 000.
Output
The output should contain a single integer — the maximal score that can be achieved.
Examples
这里写图片描述

题意：给你格子的数目和每一次能放的小格子的大小，每放一个都能得到1分，每次可以拿掉某个小格子放上新的格子，问能得到的最高分。

可以知道大小为1时是最划算的，所以放不下的时候，如果有大小为2的格子就拿掉它。

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <string>
using namespace std;
int main (void)
{
 freopen("energy.in","r",stdin);
 freopen("energy.out","w",stdout);
    string s;
    int n;
    cin>>n;
    cin>>s;
    long long ans=0,a1=0,a2=0;
    for(int i=0;i<s.size();i++)
    {
        if(s[i]=='1')
        {
            if(a1+a2*2+1<=n)    
            a1++;
            else if(a2)
            {
            a2--;
            a1++;
            }
        }
        if(s[i]=='2')
        {
            if(a1+a2*2+2<=n)
                a2++;
        }
        ans+=(a1+a2);
    }
    printf("%lld\n",ans);
    return 0;
}