Pagini recente » Cod sursa (job #852479) | Cod sursa (job #492208) | Solutia problemei shoturi | Cod sursa (job #505928) | Cod sursa (job #2280244)
#include <iostream>
#include <fstream>
#define NMAX 100001
using namespace std;
ifstream in("oo.in");
ofstream out("oo.out");
int n,maxim;
int p[NMAX+1],profit[NMAX+1];
int main()
{
in>>n;
for (int i=1;i<n;++i)
in>>p[i];
p[n+1]=p[1];
for (int i=1;i<=n+1;++i)
profit[i]=0;
for (int i=1;i<n;++i)
{
if(i-3>-1)
profit[i]=max(profit[i-1],p[i]+p[i-1]+profit[i-3]);
else
profit[i]=max(profit[i-1],p[i]+p[i-1]);
}
maxim=max(profit[n-2],maxim);
for (int i=1;i<=n+1;++i)
profit[i]=0;
for (int i=2;i<=n;++i)
{
if(i-3>0)
profit[i]=max(profit[i-1],p[i]+p[i-1]+profit[i-3]);
else
profit[i]=max(profit[i-1],p[i]+p[i-1]);
}
maxim=max(profit[n-1],maxim);
for (int i=1;i<=n+1;++i)
profit[i]=0;
for (int i=3;i<=n+1;++i)
{
if(i-3>1)
profit[i]=max(profit[i-1],p[i]+p[i-1]+profit[i-3]);
else
profit[i]=max(profit[i-1],p[i]+p[i-1]);
}
maxim=max(profit[n],maxim);
out<<maxim;
return 0;
}