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 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

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 1000005

ll 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;

}