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=1;
        y=0;
    }
    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!=0)return -1;
     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-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));
         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 0;
}

 

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