One Person Game(zoj3593+扩展欧几里德)
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;
}