Gym 100269E-Energy Tycoon

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.

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

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;}