ZOJ

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 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

20 1 1 20 1 2 4

Sample Output

1-1

题意：从A点到B点 有3种移动方式 移动 a距离，b距离或者 a+b距离，问最少需要多少步。

分析：

有3种情况

ax+by=A-B

ax+cy=A-B

bx+cy=A-B

用扩展欧几里德很容易求出 x和y的值

X=x+b/gcd *t

Y=y-a/gcd *t

通过这两个方程可以得到两个关于t的线，两个的斜率互异

我们要求的是 |X|+|Y|的最小值

所以最小值应该在 X=0或者Y=0处可以取到

但是这里的X和Y是经过整除运算得到的，所以存在误差

在计算的时候需要多查看左一位和右一位

AC代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;long long kzgcd(long long a,long long b,long long &x,long long &y){if(b==0){x=1,y=0;return a;}long long ans=kzgcd(b,a%b,x,y);long long t=x;x=y;y=t-(a/b)*y;return ans;}long long solve(long long a,long long b,long long c){long long x,y;long long gcd=kzgcd(a,b,x,y);if(c%gcd) return -1;x=x*c/gcd;y=y*c/gcd;long long aa=a/gcd;long long bb=b/gcd;//x=x0+b/gcd *t   y=y0-a/gcd *t;long long ans=abs(x)+abs(y);for(long long i=-x/bb-1;i<=-x/bb+1;i++){long long xx=x+bb*i;long long yy=y-aa*i;if(ans>abs(xx)+abs(yy))ans=abs(xx)+abs(yy);}for(long long i=y/aa-1;i<=y/aa+1;i++){long long xx=x+bb*i;long long yy=y-aa*i;if(ans>abs(xx)+abs(yy))ans=abs(xx)+abs(yy);}return ans;}int main(){int T;scanf("%d",&T);while(T--){long long A,B,a,b;scanf("%lld%lld%lld%lld",&A,&B,&a,&b);long long C=abs(A-B),c=a+b,ans[4];ans[1]=solve(a,b,C);ans[2]=solve(a,c,C);ans[3]=solve(b,c,C);if(ans[1]==-1&&ans[2]==-1&&ans[3]==-1)printf("-1\n");else{long long anss=ans[1];for(int i=2;i<=3;i++)anss=min(anss,ans[i]);printf("%lld\n",anss);}} }