#include<bits/stdc++.h>
using namespace std;
int n,a[1005],yd,nd,d,dj;//yd==yes day==打卡连续天数,nd==no day==连续遗漏天数,d==dian==点数==活跃值,dj==dian(shu)(zeng)jia==点数增加==每天增加的活跃值 
int main()
{
	cin >>n;
	for (int i=1;i<=n;i++) 
	{
		cin >>a[i];
		if (a[i]==1) 
		{
			if (a[i-1]==0) 
			{ 
				if (yd-pow(2,nd-1)>0) yd-=pow(2,nd-1);
				else yd=0;
				nd=0;
			}
			yd++;
			switch(yd) 
			{
				case 1:dj=1;
				case 3:dj=2;
				case 7:dj=3;
				case 30:dj=4;
				case 120:dj=5;
				case 365:dj=6;
			}
			d+=dj;
			
		}
		else {nd++;yd=0;}
	}
	cout <<d<<endl;
	return 0;
}
  

样例48 QwQ