One Person Game(扩展欧几里得)

One Person Game

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

 

 

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Author: CAO, Peng
Contest: The 12th Zhejiang University Programming Contest

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
ll t;
ll d1,d2,d3,d4;
ll X,Y,a,b,c;
ll GDC;

ll gdc(ll x,ll y){
    return y==0?x:gdc(y,x%y);
}

void extend_gdc(ll a,ll b,ll &X,ll &Y){
    if(b==0){
        X=1;
        Y=0;
        return;
    }
    extend_gdc(b,a%b,X,Y);
    ll temp=X;
    X=Y;
    Y=temp-a/b*Y;
}

int main(){
    ios::sync_with_stdio(false);
    while(cin>>t){
        while(t--){
            cin>>d1>>d2>>d3>>d4;
            a=d3,b=d4,c=d2-d1;
            GDC=gdc(a,b);
            if(c%GDC!=0){
                cout << "-1" << endl;
                continue;
            }
            extend_gdc(a,b,X,Y);
            X=X*c/GDC;
            Y=Y*c/GDC;
            a=a/GDC;
            b=b/GDC;
            ll mid=(Y-X)/(a+b);
            ll res=0x3f3f3f3f3f3f3f3f;
            for(int i=mid-1;i<=mid+1;i++){
                ll ans=0;
                ll ee=X+b*i;
                ll rr=Y-a*i;
                if(ee*rr>=0) ans+=max(abs(ee),abs(rr));   //同号取最大值
                else ans+=abs(ee)+abs(rr);   //异号模相加
//                cout << ans << endl;
                res=min(res,ans);
            }
            cout << res << endl;
        }
    }
    return 0;
}