One person Game (expand Euclid to find the minimum number of steps)

One person Game Time limit: 2 Seconds Memory Limit: 65536 KB

There is a interesting and simple one person game. Suppose there is a number axis under your feet. You were at point A at first and your aim was point B. There is 6 kinds of operations you can perform in one step. That's to go left or right by a,b and C, here c all equals to a+b.

You must arrive B as soon as possible. Please calculate the minimum number of steps.

Input

There is multiple test cases. The first line of input was an integer T(0 < T ≤1000) indicates the number of test cases. Then T test Cases follow. 4 integers a, B, a and b c6>, 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

20 1 1 20 1 2 4

Sample Output

1-1

Problem: It is not difficult to list the linear equation A (x+z) +b (y+z) =b-a, namely Ax+by=c;

Mainly in the middle to find the minimum number of steps;// since a+b can be substituted with C, so when X and Y are the same number, the Z=min (x, y) +max (x, y)-min (x, y) =max can be used to walk, That is, a step number can be equal to a+b // so the number of steps to find this situation ... a+b//Because the closer the x and Y, the more the *step, the more optimized, the PNS // so to be found around the point where the general solution intersects; Because the intersection may be a decimal, it is looked around;

Code:

1#include <iostream>2#include <algorithm>3#include <cstdio>4#include <cstring>5#include <cmath>6#include <vector>7 #defineMem (x, y) memset (x,y,sizeof (x))8 using namespacestd;9typedefLong LongLL;Ten Const intinf=0x3fffffff; One voidE_GCD (LL a,ll b,ll &d,ll &x,ll &y) { A     if(!b) { -D=A; -x=1; they=0; -     } -     Else{ -E_GCD (b,a%b,d,x,y); +LL temp=x; -x=y; +y=temp-a/b*y; A     } at } - ll Cal (ll A,ll b,ll c) { - LL x,y,d; - E_GCD (a,b,d,x,y); -      //printf ("%lld%lld%lld\n", d,x,y); -      if(c%d!=0)return-1; inx*=c/D; -y*=c/D; to      //x=x0+b/d*t;y=y0-a/d*t; +B/=d;a/=D; -      //since a+b can be substituted with C, so when X and Y are the same number, the      //You can use Z=min (x, y) +max (x, y)-min (x, y) =max (x, y) to walk, that is, a step number can be equal to A+b *      //so we have to find the number of steps in this situation ...  $      //because the closer the X and Y, the more The (a+b) *step, the more optimized,Panax Notoginseng      //so look around at the point where the general solution intersects, because the intersection may be a decimal, so it's looking around . -LL mid= (y-x)/(A+B);//X0+bt=y0-at; thell ans= (LL) inf*(LL) INF; + LL temp; A     //printf ("%lld\n", ans); the       for(LL t=mid-1; t<=mid+1; t++){ +          if(ABS (X+B*T) +abs (y-a*t) ==abs (x+b*t+y-a*t)) -Temp=max (ABS (X+B*T), ABS (y-a*t)); $          ElseTemp=abs (x+b*t) +abs (y-a*t); $ans=min (ans,temp); -      //printf ("%lld\n", temp); -      } the      returnans; - }Wuyi intMain () { the LL t,a,b,a,b; -scanf"%lld",&T); Wu      while(t--){ -scanf"%lld%lld%lld%lld",&a,&b,&a,&b); AboutLL ans=cal (a,b,b-A); $printf"%lld\n", ans); -     } -     return 0; -}

One person Game (expand Euclid to find the minimum number of steps)