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
2 0 1 1 2 0 1 2 4
Sample Output
1 -1
题解：不难列出线性方程a(x+z)+b(y+z)=B-A;即ax+by=C;
主要是中间找最小步数；//由于a+b可以用c来代替；所以，当x和y同号时， 34 //就可以用z=min(x,y)+max(x,y)-min(x,y)=max(x,y)来走,也就是一部分步数可以等于a+b 35 //所以还要找这种情况的步数。。。 36 //因为x和y越接近，（a+b）*step的越多，越优化， 37 //所以要在通解相交的点周围找；由于交点可能为小数，所以才在周围找的；
代码：
1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<vector> 7 #define mem(x,y) memset(x,y,sizeof(x)) 8 using namespace std; 9 typedef long long LL; 10 const int INF=0x3fffffff; 11 void e_gcd(LL a,LL b,LL &d,LL &x,LL &y){ 12 if(!b){ 13 d=a; 14 x=1; 15 y=0; 16 } 17 else{ 18 e_gcd(b,a%b,d,x,y); 19 LL temp=x; 20 x=y; 21 y=temp-a/b*y; 22 } 23 } 24 LL cal(LL a,LL b,LL c){ 25 LL x,y,d; 26 e_gcd(a,b,d,x,y); 27 //printf("%lld %lld %lld ",d,x,y); 28 if(c%d!=0)return -1; 29 x*=c/d; 30 y*=c/d; 31 //x=x0+b/d*t;y=y0-a/d*t; 32 b/=d;a/=d; 33 //由于a+b可以用c来代替；所以，当x和y同号时， 34 //就可以用z=min(x,y)+max(x,y)-min(x,y)=max(x,y)来走,也就是一部分步数可以等于a+b 35 //所以还要找这种情况的步数。。。 36 //因为x和y越接近，（a+b）*step的越多，越优化， 37 //所以要在通解相交的点周围找；由于交点可能为小数，所以才在周围找的； 38 LL mid=(y-x)/(a+b); // x0+bt=y0-at; 39 LL ans=(LL)INF*(LL)INF; 40 LL temp; 41 // printf("%lld ",ans); 42 for(LL t=mid-1;t<=mid+1;t++){ 43 if(abs(x+b*t)+abs(y-a*t)==abs(x+b*t+y-a*t)) 44 temp=max(abs(x+b*t),abs(y-a*t)); 45 else temp=abs(x+b*t)+abs(y-a*t); 46 ans=min(ans,temp); 47 // printf("%lld ",temp); 48 } 49 return ans; 50 } 51 int main(){ 52 LL T,A,B,a,b; 53 scanf("%lld",&T); 54 while(T--){ 55 scanf("%lld%lld%lld%lld",&A,&B,&a,&b); 56 LL ans=cal(a,b,B-A); 57 printf("%lld ",ans); 58 } 59 return 0; 60 }