One Person Game(zoj3593+扩展欧几里德)

One Person Game

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld
& %llu

Submit Status Practice ZOJ
3593

Appoint description: 
System Crawler  (2015-04-29)

Description

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

题意：一维坐标轴，有A和B两个地方，现在从A到B，每次可以向任意方向走a、b或者c的距离，其中c=a+b，问能不能走到B，能的话最少走几次。

思路：c = a+b, C = A-B，则等价于求{|x|+|y| | ax+by=C || ax+cy=C || bx+cy=C}。对于ax+by=C，

用扩展欧几里得算法求得ax+by=gcd(a,b)，是否有解可由C是否为gcd的倍数判断。

若有解，原方程的一组解为(x0, y0)  = (xx*C/gcd, yy*C/gcd)。

令am=a/gcd，bm=b/gcd，则通解可表示为(x0+k*bm, y0-k*am)。

能使|x|+|y|最小的整点一定是最靠近直线与坐标轴交点的地方，

可以由|x|+|y|=C的图像不断平移看出。

由于负数取整时是先对它的绝对值取整，再添上负号，

所以考虑的范围要是[-x0/bm-1, -x0/bm+1] 、[y0/am-1, y0/am+1]。

转载请注明出处：http://blog.csdn.net/u010579068/article/details/45487153

#include<stdio.h>
#include<math.h>
#include<algorithm>
#define LL long long
using namespace std;

LL ans;
void exgcd(LL a,LL b,LL& d,LL& x,LL& y)
{
    if(!b){d=a;x=1;y=0;}
    else
    {
        exgcd(b,a%b,d,y,x);
        y-=x*(a/b);
    }
}
LL China(LL a,LL b,LL c)
{
    LL x,y,d,bm,am;

    exgcd(a,b,d,x,y);
    if(c%d) return -1;
    bm=b/d;
    am=a/d;
    x=x*c/d;
    y=y*c/d;

 //   printf("x=%lld,y=%lld\n",x,y);
    LL sum=fabs(x)+fabs(y);

    for(int i=-x/bm-1;i<=-x/bm+1;i++)
    {
        LL X=x+bm*i;
        LL Y=y-am*i;
        if(i)
        {
            LL tmp=fabs(X)+fabs(Y);
            if(tmp<sum) sum=tmp;
        }
    }
    for(int i=y/am-1;i<=y/am+1;i++)
    {
        LL X=x+bm*i;
        LL Y=y-am*i;
        if(i)
        {
            LL tmp=fabs(X)+fabs(Y);
            if(tmp<sum) sum=tmp;
        }
    }
    return sum;
}

int main()
{
    int T;
    LL A,B,C,a,b,c,d,x,y;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld%lld%lld",&A,&B,&a,&b);
        c=a+b;
        C=fabs(A-B);
 //       ans==1<<30;
        LL t1=China(a,b,C);
        LL t2=China(a,c,C);
        LL t3=China(b,c,C);

        if(t1==-1&&t2==-1&&t3==-1)
        {
            printf("-1\n");
            continue;
        }
        if(t1>t2) ans=t2;
        else ans=t1;

        if(ans>t3) ans=t3;

        printf("%lld\n",ans);
    }
    return 0;
}