ZOJ 3593.One Person Game【扩展欧几里得+逼近】【4月11】

One Person Game Time Limit: 2 Seconds       Memory Limit: 65536 KB

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

根据题意，很容易知道要算 a*x+b*y = B-A中的x，y。但是解出来是一组特解，需要让x，y向0逼近

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long LL;
const LL INF = 0x3f3f3f3f;
LL e_gcd(LL a, LL b, LL &x, LL &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    LL r= e_gcd(b, a%b, x, y);
    LL temp = x;
    x = y;
    y = temp - a/b*y;
    return r;
}
int main()
{
    LL T, A, B, a, b;
    scanf("%lld", &T);
    while(T--)
    {
        LL x, y, m;
        scanf("%lld %lld %lld %lld", &A, &B, &a, &b);
        LL GCD = e_gcd(a, b, x, y);
        if((B-A)%GCD != 0) cout << -1 << endl;
        else
        {
            a /= GCD;
            b /= GCD;
            x = (B-A)/GCD*x;
            y = (B-A)/GCD*y;
            LL ans = INF*INF;
            LL t = (y-x)/(a+b);
            for(LL i = t-1;i <= t+1; ++i)
            {
                if((abs(x + b*i) + abs(y - a*i)) == abs(x + b*i + y - a*i))//同号
                {
                    m = max(abs(x+b*i), abs(y-a*i));
                }
                else m = abs(x+b*i) + abs(y-a*i);//异号
                ans = min(ans, m);
            }
            cout << ans << endl;
        }
    }
    return 0;
}