mod
krystian: jak pokazac,ze
(b+(a mod b))<= 2/3 (a+b)
gdy
a>=b
?
16 paź 01:34
wredulus_pospolitus:
dla a≥2b:
| | 2 | | 2 | |
b + (a (mod b)) ≤ b + b = |
| *(2b+b) ≤ |
| (a+b) |
| | 3 | | 3 | |
dla a=b
| | 4 | | 2 | | 2 | |
b+ (a (mod b)) = b ≤ |
| b = |
| (b+b) = |
| (a+b) |
| | 3 | | 3 | | 3 | |
dla a∊(b;2b)
niech a = 2b − k ; gdzie k∊(0,b)
L = b + (a (mod b)) = b + b−k
| | 2 | | 2 | |
P = |
| (a+b) = |
| (2b−k + b) |
| | 3 | | 3 | |
3b+3b−3k ≤ 4b+2b−2k
0 ≤ k
c.n.w.
16 paź 08:45