ZOJ

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,band 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

20 1 1 20 1 2 4

Sample Output

1-1

题意：给你一个起点和终点，每次可以向左或向右走a步或b或c步，c=a+b;问最小步数；

根据公式ax+by=c;，当x,y同号时等于max（x,y）,当a,b异号时等于（abs(x)+abs(y)）,因为a,b大于0，所以不管x,y同号还是异号都是当x,y,最接近时，答案最小，写出通式

x=x1+b/(gcd)*k,y=y1-a/gcd*k; 假设x,y相等,k=(y-x)/(a+b),因为x,y不一定相等,所以枚举k附近的点

#include <algorithm>#include <string.h>#include <iostream>#include <stdio.h>#include <string>#include <vector>#include <queue>#include <map>#include <set>using namespace std;typedef long long LL;const int N = 1e5+10;LL exgcd(LL a,LL b,LL &x,LL &y){    if(b==0)    {        x=1, y=0;        return a;    }    LL ans=exgcd(b,a%b,x,y);    LL tmp=x;    x=y;    y=tmp-a/b*y;    return ans;}int main(){    int t;    scanf("%d", &t);    while(t--)    {        LL A, B, a, b, c, x, y;        scanf("%lld %lld %lld %lld",&A, &B, &a, &b);        if(A>B) swap(A,B);        c=(B-A);        LL g=exgcd(a,b, x, y);        if(c%g!=0) puts("-1");        else        {            a/=g, b/=g;            x*=(c/g), y*=(c/g);            LL k=(y-x)/(a+b);            LL ans=10000000000ll;            for(LL i=k-1;i<=k+1;i++)            {                if(abs(x+i*b)+abs(y-i*a)==abs(x+i*b+y-i*a)) ans=min(ans,max(abs(x+i*b),abs(y-i*a)));                else   ans=min(ans,abs(x+i*b)+abs(y-i*a));            }            cout<<ans<<endl;        }    }    return 0;}