What is 赛瓦维斯特 Theorem
If I just say 赛瓦维斯特 theorem, you may not know what it is (otherwise you will not come point saw); then if I say $ NOIP \ 2017 \ D1 \ T1 $ Oscar doubts that you may be understood why.
In fact, 赛瓦维斯特 Theorem is:
Known $ a, b $ greater than $ 1 $ is a positive integer, $ gcd (a, b) = 1 $, is so indeterminate equation $ ax + by = C $ non-negative solution to the largest integer $ C = a \ times bab $.
Well, that doubt Oscar in $ a \ times bab $.
赛瓦维斯特 theorem proving
Or to consider that question, the examination room to find the law of an average $ 20 $ minutes to find the law; but who want to "positive solution" of children's shoes, about $ 1 $ hours to $ A $ off of this question (essentially used We are extended Euclidean).
However, it is clear that the law is very remarkable to find, so we now consider real positive solution, that is, to prove the law.
Let's prove $ a \ times bab $ must not be taken to.
Using contradiction ( arc method ), we assume that there $ x, y \ geqslant 0 $ satisfies $ ax + by = ab-ab $.
Transposition obtained: $ A (X +. 1) + B (Y +. 1) = ab & $ ① .
We then ab & $ $ to addition to the left, i.e. $ \ frac {a (x + 1)} {ab} + \ frac {b (y + 1)} {ab} = 1 $, what can be obtained in the extinction $ \ frac {x + 1} {b} + \ frac {y + 1} {a} = 1 $.
So $ a | (y + 1), b | (x + 1) $, where $ | $ is divisible symbol (once exams shadows emphasize here ......)
Prove the simple: we return formula ①, which one then can be obtained $ b (y + 1) = a (bx-1) $, because there is $ gcd (a, b) = 1 $, Therefore $ a | (y + 1) $, $ b | (x + 1) $ Similarly.
又$\because a(x+1)+b(y+1)\geqslant a\times+b\times a=2ab$（$\because b|(x+1)\ \$therefore b\leqslant x+1$，对于$a\leqslant y+1$同理）。
We assume that in the $ a (x + 1) + b (y + 1) = ab, a \ geqslant 0, b \ geqslant 0 $ contradictory.
$ \ $ THEREFORE assumption does not hold, i.e., absence of $ x, y \ geqslant 0 $, satisfying $ ax + by = ab-ab $.
$ \ Therefore $ $ a \ times bab $ will not be taken to the original title in.
Now we have to prove that for any positive integer $ C \ geqslant ab-ab + 1 $ must be taken to.
Transposition now have the formula: $ C + a + b \ geqslant ab + 1 $.
Wish to set $ k, m $ to satisfy $ C + a + b = ka + m (k \ geqslant b, 1 \ leqslant m \ leqslant a-1) $.
And $ \ because gcd (a, b) = 1 $, may be formed Pei Shu Theorem (when $ gcd (a, b) = 1 $, then there exists $ x, y \ in Z $, so $ ax + by = $ 1) if obtained $ x ', y' \ in Z, - (b-1) \ leqslant x '\ leqslant -1 $, then there exists an integer $ y' $ such that $ ax '+ by' = m $.
A simple proof at the above sentence, $ \ because $ positive integer $ x '$ go into a total number $ b-1 $, $ y' = \ frac {m-ax '} {b} $, will be able to find the $ x '$ such that $ b | (m-ax') $.
又$\because ax'<0,m>0,b>0$，$\therefore y\geqslant 1$。
$\therefore x=k+x'-1,y=y'-1$。
又$\because x,y\geqslant 0$。
$\therefore ax+by=C$
$ \ $ THEREFORE for any $ C \ geqslant ab-ab + 1 $ must exist $ x, y \ geqslant 0 $ satisfies $ ax + by = C $.
QED.
(I hope it is validated, evidence was wrong not to consume)
example
$ \ Alpha.NOIP \ 2017 \ D1 \ T1 $ Oscar doubts (do not poke, no link).
$ \ beta. $ $ 2 $ small odd mining .
rp ++