zoj3593One Person Game (扩展欧几里德）

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 6kinds 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 testcase 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

题意：有两个端点A，B，每次你可以向左或者向右走a,b,a+b的距离，问最少走多少次能从A走到B。

思路：设A,B之间的距离为dis，a+b=c,那么题目等价于min{|x|+|y| | (ax+by=dis) || (ax+cy=dis) || (bx+cy=dis)  }.注：扩展欧几里德算出来ax+by=gcd(c,d)中的特殊值x,y一定满足|x|+|y|最小，但是如果算的是ax+by=d，(d%gcd(a,b)，但是d！=gcd(a,b))那么就不一定|x|+|y|最小，此时要使得x趋近于0，或者使得y趋近于0，这样算出来的k带入然后取几者中的最小值，因为x，y的关系式是一条直线，|x|+|y|=m在直线接近坐标轴的情况下取到最小值。

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;typedef long long ll;typedef long double ldb;#define inf 10000000000000LL#define pi acos(-1.0)#define MOD 1000000007#define maxn 1000005ll extend_gcd(ll a,ll b,ll &x,ll &y){if(b==0){x=1;y=0;return a;}ll d=extend_gcd(b,a%b,y,x);y-=a/b*x;return d;}ll niyuan(ll a,ll n){ll x,y;ll d=extend_gcd(a,n,x,y);if(d==1) return (x%n+n)%n;else return -1;}ll gcd(ll a,ll b){return (b>0)?gcd(b,a%b):a;}ll solve(ll a,ll b,ll dis){ll x,y,d,x0,y0,ans;d=extend_gcd(a,b,x,y);if(dis%d!=0)return -1;x0=x*dis/d;y0=y*dis/d;ll aa,bb;aa=a/d;bb=b/d;ll k;k=-x0/bb-1;ans=abs(x0+bb*k)+abs(y0-aa*k);k++;ans=min(ans,abs(x0+bb*k)+abs(y0-aa*k) );k++;ans=min(ans,abs(x0+bb*k)+abs(y0-aa*k) );k=y0/aa-1;ans=min(ans,abs(x0+bb*k)+abs(y0-aa*k) );k++;ans=min(ans,abs(x0+bb*k)+abs(y0-aa*k) );k++;ans=min(ans,abs(x0+bb*k)+abs(y0-aa*k) );return ans;}int main(){int n,m,i,j,T;ll x1,x2,a,b,c,x,y;scanf("%d",&T);while(T--){scanf("%lld%lld%lld%lld",&x1,&x2,&a,&b);ll dis=abs(x2-x1);c=a+b;ll ans=solve(a,b,dis);ans=min(ans,solve(a,c,dis));ans=min(ans,solve(b,c,dis));printf("%lld\n",ans);}return 0;}