//题目连接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4677//题目大意:每次可以向左或者向右走a或者b或者(a+b)步.问在数轴上从A->B最少需要多少次.//解题思路:式子ax+by=c;如果x>0&&y>0 那么ans=max(x,y);if(x<0&&y>0)ans=y-x// 所以就是去求一组解(x,y)使得上面的ans最小..// 在采用EXGCD求得一组解之后(这里注意.这里的解有可能很大.10^14).分别讨论1.x<0&&y>0// 2.x>0&&y>0 3.x>0&&y<0 .. 这三种情况的时候t存在一个范围.// (因为t范围是由分式得来.所以开始做的时候通过两边同乘来讨论t,后来发现t很大.做乘法会超过LL)// 后来直接用分式去计算t能够所在区间范围..然后在三组优解中找到一个最优解即可.#include<stdio.h>#include<stdlib.h>#include<iostream>#include<math.h>#include<vector>#include<string.h>#include<algorithm>using namespace std;#define LL long longLL GCD(LL a , LL b){ if(a > b) { LL c = a; a = b; b = c ; } while(a) { LL m = a; a = b%a; b = m; } return b;}void exGCD(LL a , LL b , LL& x , LL& y){ if(!b) { x = 1; y = 0; return ; } else exGCD(b,a%b,y,x); y -= x * (a / b);}LL judge(LL x , LL y){ // printf("%lld %lld\n",x,y); if(x >= 0 && y >= 0) { if(x > y)return x; return y; } if(x >= 0 && y <= 0)return x - y; if(x <= 0 && y >= 0)return y - x; return 0x3f3f3f3f;}LL sum(LL x , LL y ,LL p , LL q , LL t){ x = x + p * t; y = y - q * t; return judge(x,y);}bool get(LL x , LL y){ if(x < 0) x = - x; if(x % y == 0)return 1; return 0;}LL solve(LL a , LL b , LL c){ LL x , y; LL ans = 1e18; LL p = GCD(a,b); if(c % p != 0)return -1; exGCD(a,b,x,y); x *= c / p ; y *= c / p; ans = min(ans , judge(x,y)); LL q = a/p; p = b/p; LL t ,t1,t2; if(x >= 0 && y <= 0) { if(get(-x , p))t1 = (-x) / p; else t1 = - (x / p); if(get(y , q))t2 = y / q; else t2 = - (-y / q); t = max(t1,t2); ans = min(ans , sum(x,y,p,q,t)); if(get(y , q))t2 = y / q; else t2 = - ((-y) / q + 1); t = (y-x)/(p+q); if(t1 <= t2) { if(t >= t1 && t <= t2) { ans = min(ans , sum(x,y,p,q,t)); ans = min(ans , sum(x,y,p,q,t + 1)); ans = min(ans , sum(x,y,p,q,t - 1)); } else { if(t1 > t) ans = min(ans , sum(x,y,p,q,t1)); else ans = min(ans , sum(x,y,p,q,t2)); } } if(get(-x , p))t1 = (-x) / p; else t1 = - (x / p + 1); t = min(t1,t2); ans = min(ans , sum(x,y,p,q,t)); } else { if(get(-x,p))t1 = -x / p; else t1 = (-x) / p; if(get(y,q))t2 = y / q; else t2 = y / q; t = min(t1,t2); ans = min(ans , sum(x,y,p,q,t)); if(get(-x,p))t1 = -x / p; else t1 = (-x) / p + 1; t = (y - x) / (p + q); if(t1 <= t2) { if(t >= t1 && t2 >= t) { ans = min(ans , sum(x,y,p,q,t)); ans = min(ans , sum(x,y,p,q,t-1)); ans = min(ans , sum(x,y,p,q,t+1)); } else { if(t < t1)ans = min(ans , sum(x,y,p,q,t1)); else ans = min(ans , sum(x,y,p,q,t2)); } } if(get(-x,p))t1 = (-x) / p; else t1 = (-x) / p + 1 ; t = max(t1,t2); ans = min(ans , sum(x,y,p,q,t)); } return ans;}int main(){ int test; // freopen("in.txt","r",stdin); //freopen("out3.txt","w",stdout); scanf("%d",&test); while(test--) { LL a,b,A,B; scanf("%lld%lld%lld%lld",&A,&B,&a,&b); if(A > B) { LL p = A; A = B; B = p; } LL c = B - A; printf("%lld\n",solve(a,b,c)); }}
