扩展欧几里得exgcd

1.https://www.luogu.org/problemnew/solution/P1082
2.求 (ax + by = gcd(a,b))中的未知数
3.(由axequiv1(mod b))可得ax+by=1
a，b为gcd(a,b)倍数,so (ax+by=z（if为1）)要有解，z（if为1）必须为gcd(a,b)（那 么为一）的倍数
4.再根据gcd反推
5.code

x的返回值必须为1=》gcd(a,b)x+0*y=gcd(a,b);
y的返回值无要求，但要小心溢出
(ax+by=bx_2+(a mod b)y_2)
(ax+by=ay_2+b(x_2−(a/b)y_2))
​

#include<bits/stdc++.h>
#define re return
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define ll long long 

char buf[1<<21],*p1,*p2;
inline int getc(){re p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}

template<typename T>inline void rd(T&x)
{
    char c;bool f=0;
    while((c=getc())<'0'||c>'9')if(c=='-')f=1;
    x=c^48;
    while((c=getc())>='0'&&c<='9')x=x*10+(c^48);
    if(f)x=-x;
}


using namespace std;
ll a,b,c,x,y;

ll exgcd(ll a,ll b)
{
    if(b==0){x=1;y=0;re a;}
    ll ans=exgcd(b,a%b);
    ll k=x;
    x=y;
    y=k-a/b*y;
    re ans;
}

int main()
{
//	freopen("in.txt","r",stdin);
    rd(a),rd(b);
    int z=exgcd(a,b);
    printf("%lld",(x%b+b)%b);
    re 0;
}