51nod 1256 乘法逆元

#include <iostream>

using namespace std;

int a,b,x,y;

int exGcd(int a, int b, int& x, int& y)
{
if (b == 0)
{
x = 1;
y = 0;
return a;
}
int g = exGcd(b,a%b,x,y);
int tmp = x;
x = y;
y = tmp - (a / b) * y;
return a;
}

int main()
{
std::ios::sync_with_stdio(false);
cin >> a >> b;
exGcd(a,b,x,y);
cout << (x%b+b)%b << endl;
return 0;
}