zoj 3593 One Person Game

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4677

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 toa+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

 题目大意：一个人的起点是A，终点是B，他每走一步可以选择走a米，也可以走b米，还可以走a+b米，可以往左走也可以往右走，问从A走到B最少需要走几步

可以设走x步a米的，走y部b米的，得到式子：

ax + by = B - A;

通过扩展欧几里德可得x和y

x、y的解集：X = x + b/gcd(a,b) * t;

                  Y = y - a /gcd(a,b) * t;

可将X、Y的式子看做两个直线方程，她们一个斜率为正一个斜率为负，肯定会有交点t，交点t处X==Y，此时所走的步数最少，但步数一定为整数，如果t为小数，就不能取t，需要往t的左右两边找（t - 1， t + 1）这个范围内肯定存在一个整数t使得所走步数最少，根据t可求得X、Y

（1）X、Y同号时，即方向相同，可以走a+b来减少步数，取max(X,Y)（这里可以画X、Y的坐标图来帮助理解）

（2）X、Y异号时，即方向相反，步数绝对值相加，abs(X) + abs(Y)

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>

using namespace std;

typedef long long ll;

const ll INF = 0x3f3f3f3f;

ll r;

void gcd(ll a, ll b, ll &x, ll &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        r = a;
        return ;
    }
    gcd(b, a % b, x, y);
    ll t = x;
    x = y;
    y = t - a / b * y;
}

ll solve(ll a, ll b, ll c)
{
    ll x, y, m;
    gcd(a, b, x, y);
    if(c % r != 0)
        return -1;
    a /= r;
    b /= r;
    x = c / r * x;
    y = c / r * 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))//x、y同号
            m = max(abs(x + b * i), abs(y - a * i));
        else
            m = abs(x + b * i) + abs(y - a * i);
        ans = min(ans, m);
    }
    return ans;
}

int main()
{
    int t;
    ll A, B, a, b, c;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%lld%lld%lld%lld", &A, &B, &a, &b);
        c = B - A;
        ll ans = solve(a, b, c);
        printf("%lld
", ans);
    }
    return 0;
}