NOIp D1T1 Oscar doubts

Make complaints

Sure enough, people are wondering this question, for it is my kind of math slag slag too unfriendly, and where the conclusions want, guess or guess.

A thought

Pure mathematics, seen quickly cut off, have not seen it ...

The conclusion is: known $ a, b $ is a positive integer greater than $ 1 $ coprime, the so indefinite equation $ A X + B Y = C $ not nonnegative integer Solutions maximum integer $ C = A B - A - b $

What seems to be called赛瓦维斯特theorems, but in addition to the solution to a problem of solving the problem, I do not have to search anywhere else, associated with mathematical signs not at all, amazing, is it a mere $ OIer $ before the study formula ?

Prove it.

First, the proband $ A X + B Y = A B - A - B (A, B>. 1, (A, B) =. 1) $ nonnegative integer solution exists.

By contradiction, assuming the presence of $ x> = 0, y> = 0 $, so $ A X + B Y = A B - A - B (A, B>. 1, (A, B) =. 1) $ established.

Transposition to give $ a * (x + 1) + b * (y + 1) = a * b $

$a*(x+1)=b*(a-y-1)$

And because $ (a, b) = 1 $

Then $ b | (x + 1) $

Same reason: $ a | (y + 1) $

And because $ x> = 0, y> = 0 $

Therefore, $ x> = b, y> = a $

则$a*(x+1)+b*(y+1)>=ab+ba>=2ab$

Because $ a> 1, b> 1 $

So $ ab> 1 $

So $ 2ab> ab $

Previously assumed $ a * (x + 1) + b * (y + 1) = a * b $ contradiction, so the assumption is not valid.

 

Next, to prove $ A X + B Y = C (A, B>. 1, (A, B) =. 1) $, for all $ C> ab-ab $, equations are non-negative integers Solutions

设$C=ka+m-a-b(k>=b,a<=m<=a-1)$,即$ax+by=ka+m-a-b(k>=b,1<=m<=a-1)$

Because $ (a, b) = 1 $, according to the theorem of Shu Pei, found the presence of $ x_0 , y_0∈Z $, so $ ax_0 + by_0 = 1 $

There x_0 $ , $ y_0∈Z, so $ ax_0 + by_0 = m $

$ Y_0 = (m-ax_0) / b $, for different $ m% b $, there are $ b-1 $ $ x_0 $ kind of value, such $ $ integer y_0

We made $ - (b-1) <= x_0 <= - 1 $, in order to guarantee the first $ y_0> = 0 $

Because $ -ax_0> 1, m> = 1 $, so in fact $ y_0> = 1 $

Then take $ y = y_0-1 $, the $ y> = 0 $

则$x_0=(m-by_0)/a$,

$x=(ka+m-a-b-by)/a=k-1+(m-b-by)/a=k-1+(m-b-b(y0-1))/a=k-1+(m-by_0)/a=k-1+x_0$

And because $ - (b-1) <= x_0 <= - 1 $, the $ - (b-1) + k-1 <= x <= - 1 + k-1 $, $ - b + k <= x <= k-2 $

And because $ k> = b $, the $ -b + k> = 0 $, the $ x> = 0 $

Proved.

Ideas two

Violence hit the table to find the law, but under no circumstances OEIS I usually not to be found ...

 

 

 

 

 

 

 

That depends on luck

Code

To be honest, this question can not paste the code qwq

 

 1 #include<iostream>
 2 #include<string>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<queue>
 6 #include<algorithm>
 7 #include<vector>
 8 using namespace std;
 9 #define N 255
10 #define ll long long 
11 #define INF 0x3f3f3f3f
12 ll a,b;
13 int main()
14 {
15     scanf("%lld %lld",&a,&b);
16     printf("%lld\n",a*b-a-b);
17     return 0;
18 }