in pairs, no?

He missteps on a bomb.

Which is 19, not 17. That's why we are asking how can it be 17?

Finishing!
Let we have 1 10 20 25 30
We are begin with star: (1,10),(1,20),(1,25),(1,30);
but 2*(1,10)+(30,30) < (1,10)+(1,30)+(1,30) because 2*10 < 30+1  (i.e. 2*a[1]<a[k]+a[0])
We form star+matching: 2*(1,10),(1,20) +(25,30)-optimal;
Let now we have 1 5 12 12 14 14 16 19
In this situation 2*5<12+1 and star converts to matching at hole:
4*(1,5)+(12,12),(14,14),(16,19).
For 1 10 10 10 10 10 15  18 19  we have  2*10>18+1
and star (1,10),(1,10),(1,10),(1,10),(1,15),(1,18),(1,19) is optimal

AC!! Not DP Not Greedy and any search . Just strict calculation of solution.
Many submissiond used because greedy more simple to program
than structural logics.


Edited by author 13.10.2011 12:23

Edited by author 13.10.2011 12:31

Right answer is 13

OK, what about:
6
1 20 30 30 2011 2011 ?

Me too.