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

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：寻找&星空の孩子

 #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=;y=;}
     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 -;
     bm=b/d;
     am=a/d;
     x=x*c/d;
     y=y*c/d;

     LL sum=fabs(x)+fabs(y);

     for(int i=-x/bm-;i<=-x/bm+;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-;i<=y/am+;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);

         LL t1=China(a,b,C);
         LL t2=China(a,c,C);
         LL t3=China(b,c,C);

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

         if(ans>t3) ans=t3;

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