Extended Euclidean algorithm--exgcd (P1082 congruence equation)

exgcd is used to find $a\ast x+b\ast and=gcd(a,b)$ a set of solutions{x, y} \{x, y\}{ x,y } method. $$a\ast x+b\ast and=gcd(a,b)---1$$$$b\ast x+(a\%b)\ast and=gcd(b,a\%b)---2$$$$a\%b=a-[a/b]\ast b---3$$$$Thethreebandinto2=>$$$$b\ast x+(a-[a/b]\ast b)\ast and=gcd(b,a\%b)---4$$$$Thefourvariable-shape=>$$$$a\ast and+b\ast (x-[a/b]\ast and)=gcd(b,a\%b)---It$$ can be seen that$$5$$$$May\; betolook\; attheturnintotheoneof\; theformtype,itmay\; bein\; ahandovernormalizedseekingsolutionof.$$$$Straighttob=0o\text{'}clock,a=gcd，a\ast x+b\ast and=gcd=>$$$$Specialsolutionx=1，and=0,thenthere-transpushedmostopenstartA,Bof\; thesolution;$$

int exgcd(int a, int b, int &x, int &y) {
    
    
	if(!b) {
    
    
		x = 1, y = 0;
		return a;
	}
	int gcd = exgcd(b, a%b, x, y);
	//上面部分和求最大公因子一样。
	//下面部分就是证明中式子5的x, y变换。
	//
	int temp = x; 
	x = y;
	y = temp - a/b * y;
	return gcd;
}

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P1082 Congruence equation

Question: Given $a,p$ , find satisfying$a\ast x=1\left(modp\right)$ x;
to use the above exgcd, we first set$a\ast x=1\left(modp\right)$ becomes a form.
$a\ast x=1(modp)=a\ast x+p\ast and=1$ . In this way, we convert to finding x, y. But in exgcd$a\ast x+p\ast y$ is equal to$gcd(a,b)$ , so 1 must be divisible by$gcd(a,b)$ . This is equivalent to$gcd(a,b)=1$ , a and b are relatively prime. The question says there must be a solution, and the non-existent situation is not considered.
Because x may be less than 0,(x+m)%mmake x greater than 0.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
void exgcd(ll a, ll b, ll &x, ll &y) {
    
    
	if(!b) {
    
    
		x = 1, y = 0;
		return ;
	}
	exgcd(b, a%b, x, y);
	ll temp = x;
	x = y;
	y = temp - a/b*y;
}
int main() {
    
     
	ll n, m, x, y;
	scanf("%lld%lld", &n, &m);
	exgcd(n, m, x, y);
	printf("%lld", (x+m)%m);
	return -0;
}