One Person Game（扩展欧几里德求最小步数）

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

题解：不难列出线性方程a(x+z)+b(y+z)=B-A;即ax+by=C;

主要是中间找最小步数；//由于a+b可以用c来代替；所以，当x和y同号时， 34 //就可以用z=min(x,y)+max(x,y)-min(x,y)=max(x,y)来走,也就是一部分步数可以等于a+b 35 //所以还要找这种情况的步数。。。 36 //因为x和y越接近，（a+b）*step的越多，越优化， 37 //所以要在通解相交的点周围找；由于交点可能为小数，所以才在周围找的；

代码：

 #include<iostream>
 #include<algorithm>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<vector>
 #define mem(x,y) memset(x,y,sizeof(x))
 using namespace std;
 typedef long long LL;
 const int INF=0x3fffffff;
 void e_gcd(LL a,LL b,LL &d,LL &x,LL &y){
     if(!b){
         d=a;
         x=;
         y=;
     }
     else{
         e_gcd(b,a%b,d,x,y);
         LL temp=x;
         x=y;
         y=temp-a/b*y;
     }
 }
 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!=)return -;
      x*=c/d;
      y*=c/d;
      //x=x0+b/d*t;y=y0-a/d*t;
      b/=d;a/=d;
      //由于a+b可以用c来代替；所以，当x和y同号时，
      //就可以用z=min(x,y)+max(x,y)-min(x,y)=max(x,y)来走,也就是一部分步数可以等于a+b
      //所以还要找这种情况的步数。。。
      //因为x和y越接近，（a+b）*step的越多，越优化，
      //所以要在通解相交的点周围找；由于交点可能为小数，所以才在周围找的；
      LL mid=(y-x)/(a+b); // x0+bt=y0-at;
      LL ans=(LL)INF*(LL)INF;
      LL temp;
     // printf("%lld\n",ans);
      for(LL t=mid-;t<=mid+;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));
          else temp=abs(x+b*t)+abs(y-a*t);
          ans=min(ans,temp);
      //    printf("%lld\n",temp);
      }
      return ans;
 }
 int main(){
     LL T,A,B,a,b;
     scanf("%lld",&T);
     while(T--){
         scanf("%lld%lld%lld%lld",&A,&B,&a,&b);
         LL ans=cal(a,b,B-A);
         printf("%lld\n",ans);
     }
     return ;
 }

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