NOIP 2017 Xiaokai's doubts

Thought

1. a,b Mutual quality
2. Find the maximum number k that cannot be expressed
3. Then ka,b Mutual quality

Here is a conclusion: (provided online)But I am the law of looking for a watch.

Ifx,y(set x<y) Mutual quality:
$nx\equiv$nx≡a (mod y) if you double x, you can get

At this time, the value of a just traverses all the numbers in 0 ~ y-1.

Assume k>x*y if we want to use x, y to spell out k

k=p*y+c(0<c<y && p>=x)

Again $\because$∵ nx mod y = c and n < p , k = n * x + (p-n) * y
$\therefore$∴ k must be spelled out by x, y

It can be seen that the k we require must be <x*y

also $\because$∵ x*y $\equiv$≡y (mod y)
x(y-1) $\equiv$≡y-x (mod y)
x(y-1)-(y-1) $\equiv$≡ 1-x
(x-1)(y-1) $\equiv$≡ 1-x
(x-1)(y-1)-1 $\equiv$≡ -x
again $\because$∵ x, y is a positive integer
$\therefore$∴ -x<0, there is no solution for x, y>0
Ok, actually, let’s focus on here.
The code is as follows:

#include<iostream>
using namespace std;
int main()
{
	long long a,b;//a,b must open long long 
	cin>>a>>b;
	cout<<a*b-a-b<<endl;
	return 0;
}