nefu 630 Min Chain(扩展欧几里得)

Raven likes games of numbers. Today he meets two numbers and thinks whether he could get a result of 1 by doing at least one operation (addition or subtraction). However, he is tired of calculation; he also wants to know the minimum steps of operation that he could get 1.

The first line of the input contains an integer T, which indicates the number of test cases.
In the following T rows, there are two positive integers a, b ( 0<=a, b<=10^9) in each row.

For each case, output the least number of steps.
If you cannot get 1, just output -1.

Sample Input
3
3 2
16 9
6 8

1
10
-1

Sample 1: 3 - 2 = 1， One subtraction will be needed.
Sample 2: 16-9+16-9+16-9-9-9+16-9-9=1，It requires 10 additions and subtractions.
Sample 3: You cannot get 1.

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<stack>
#include<algorithm>
using namespace std;
long long a,b;
long long exgcd(long long m,long long &x,long long n,long long &y)
{
long long x1,y1,x0,y0;
x0=1;y0=0;
x1=0;y1=1;
long long r=(m%n+n)%n;
long long q=(m-r)/n;
x=0;y=1;
while(r)
{
x=x0-q*x1;y=y0-q*y1;x0=x1;y0=y1;
x1=x;
y1=y;
m=n; n=r;r=m%n;
q=(m-r)/n;
}
return n;
}
int main()
{
int t;
long long x,y;
cin>>t;
while(t--)
{
scanf("%d%d",&a,&b);
if(b>a)
swap(a,b);
if(b==0)
{
if(a==1)
printf("1\n");
else
printf("-1\n");
}
else
if(b==1&&a==2)
printf("1\n");
else
if(b==1)
printf("2\n");
else
{
long long d=exgcd(a,x,b,y);
//cout<<x<<"&&"<<y<<endl;
if(d!=1)
printf("-1\n");
else
{

long long c=1,e;
e=x*c/b;
x=c*x-e*b;
y=(1-a*x)/b;
long long ans=fabs(x)+fabs(y);
printf("%lld\n",ans-1);
}
}
}

return 0;
}