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  1. def f(sum1,cs,c1,c2,c3,mode)
  2. 	s3=sum1-cs[c3]+cs[c1]
  3. 	s2=cs[c3]-cs[c2]
  4. 	s1=cs[c2]-cs[c1]
  5. 	if mode==true then
  6. 		return [s3,s2,s1].min
  7. 	else
  8. 		return [s3,s2].min
  9. 	end
  10. end
  11.  
  12. cs=[0]
  13. sum=0
  14. n=gets.to_i
  15. n.times{
  16. 	sum+=gets.to_i
  17. 	cs<<sum
  18. }
  19. c1=0
  20. c2=1
  21. ans=0
  22.  
  23.  
  24. c1=0
  25. c2=1
  26. c3=2
  27. while c1<n-2
  28. 	c2=c1+1 if c2<=c1
  29. 	while c2<n-1
  30. 		c3=c2+1 if c3<=c2
  31. 		while c3<n
  32. 			next if c3+1==n && c1==0
  33. 			s3L=f(sum,cs,c1,c2,c3+1,false)
  34. 			s3R=f(sum,cs,c1,c2,c3,false)
  35. 			#p [c1,c2,c3,s3L,s3R]
  36. 			if s3L<=s3R then
  37. 				ans=[ans,f(sum,cs,c1,c2,c3+1,false)].max
  38. 				c3+=1
  39. 			else
  40. 				ans=[ans,f(sum,cs,c1,c2,c3,false)].max
  41. 				break
  42. 			end
  43. 		end
  44. 		s2L=f(sum,cs,c1,c2,c3,true)
  45. 		s2R=f(sum,cs,c1,c2+1,c3,true)
  46. 		if s2L<=s2R	
  47. 			c2+=1
  48. 		else
  49. 			break
  50. 		end
  51. 	end
  52. 	c1+=1
  53. end
  54. puts ans
Success #stdin #stdout 0.01s 7928KB
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stdout
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981
https://ideone.com/nM6bT0
language:
Ruby (ruby 2.5.5)
created:
2 days ago
visibility:
public

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