题解：One Person Game（扩展gcd,求|X|+|Y|最小值）

There is an interesting and simple one person game. Suppose there is a number axis under your feet. You are at point A at first and your aim is point B. There are 6 kinds of operations you can perform in one step. That is to go left or right by a,b and c, here c always equals to a+b.
You must arrive B as soon as possible. Please calculate the minimum number of steps.

Input

There are multiple test cases. The first line of input is an integer T(0 < T ≤ 1000) indicates the number of test cases. Then T test cases follow. Each test case is represented by a line containing four integers 4 integers A, B, a and b, separated by spaces. (-231 ≤ A, B < 231, 0 < a, b < 231)

Output
For each test case, output the minimum number of steps. If it’s impossible to reach point B, output “-1” instead.

    Sample Input
     
    
    2
    0 1 1 2
    0 1 2 4

     
    

    Sample Output
     
    
    1
   -1

求
满足ax+by=|B-A|
|x|+|y|最小值

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f;
void exgcd(ll a,ll b,ll&g,ll&x,ll&y)//g:gcd(a,b)
{
 if(b==0)
 {
  x=1;y=0;g=a;return;
 }
 exgcd(b,a%b,g,y,x);
 y-=(a/b)*x;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        ll A,B,C,a,b;
        cin>>A>>B>>a>>b;
        ll x,y;
        ll r;
        exgcd(a,b,r,x,y);
        C=B-A;
        if(C%r)
            cout<<"-1"<<endl;
        else
        {
            x*=(C/r);
            y*=(C/r);
            a/=r;
            b/=r;
            ll ans=inf*inf,tmp;
            ll mid=(y-x)/(a + b);//x|+|y|最小值在x=x+bt与y=y-at两条直线交点附近；
            for(ll T=mid-1;T<=(mid+1);T++)
            {
                if((x+b*T)*(y-a*T)>=0)//同方向 ，可合并 
                {
                    tmp=max(abs(x+b*T),abs(y-a*T));
                }
                else// 
                    tmp=abs(x-y+(a+b)*T);
                ans=min(ans,tmp);
            }
            cout<<ans<<'\n';
        }
    }
    return 0;
}