One Person Game (zoj3593 + Extended Euclidean), personzoj3593

One Person Game (zoj3593 + Extended Euclidean), personzoj3593
One Person GameTime Limit:2000 MSMemory Limit:65536KB64bit IO Format:% Lld & % maid 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 pointAAt first and your aim is pointB. There are 6 kinds of operations you can perform in one step. That is to go left or rightA,BAndC, HereCAlways equalsA+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 integerT(0 <T≤ 1000) indicates the number of test cases. ThenTTest cases follow. Each test case is represented by a line containing four integers 4 integersA,B,AAndB, Separated by spaces. (-231 ≤A,B& Lt; 231, 0 & lt;A,B<231)

Output

For each test case, output the minimum number of steps. If it's impossible to reach pointB, Output "-1" instead.

Sample Input

20 1 1 20 1 2 4

 

Sample Output

1-1

 

 

 

 

 

The one-dimensional coordinate axis has two locations: A and B. from A to B, the distance between a, B, and c can be taken in any direction at a time. c = A + B, ask if I can go to B. If I can, I can go to B at least a few times. Idea: c = a + B, C = A-B, it is equivalent to {| x | + | y | ax + by = C | ax + cy = C | bx + cy = C }. For ax + by = C, use the Extended Euclidean Algorithm to obtain ax + by = gcd (a, B). Whether the solution exists can be determined by whether C is a multiple of gcd. If there is a solution, the group of the original equation is (x0, y0) = (xx * C/gcd, yy * C/gcd ). So am = a/gcd, bm = B/gcd, then the general solution can be expressed as (x0 + k * bm, y0-k * am ). Can make | x | + | y | the smallest entire point must be the closest to the intersection of the straight line and the coordinate axis, which can be seen by the continuous translation of the | x | + | y | = C image. Since when a negative number is rounded up, the absolute value of a negative number is rounded up, and then the negative number is added, so the scope of consideration is [-x0/bm-1, -x0/bm + 1], [y0/am-1, y0/am + 1].

 

Reprinted, please specify the Source: search for & STAR kids

 

1 # include <stdio. h> 2 # include <math. h> 3 # include <algorithm> 4 # define LL long 5 using namespace std; 6 7 LL ans; 8 void exgcd (LL a, LL B, LL & d, LL & x, LL & y) 9 {10 if (! B) {d = a; x = 1; y = 0;} 11 else12 {13 exgcd (B, a % B, d, y, x ); 14 y-= x * (a/B); 15} 16} 17 LL China (LL a, LL B, LL c) 18 {19 LL x, y, d, bm, am; 20 21 exgcd (a, B, d, x, y); 22 if (c % d) return-1; 23 bm = B/d; 24 am = a/d; 25 x = x * c/d; 26 y = y * c/d; 27 28 29 LL sum = fabs (x) + fabs (y); 30 31 for (int I =-x/bm-1; I <=-x/bm + 1; I ++) 32 {33 ll x = x + bm * I; 34 ll y = y-am * I; 35 if (I) 36 {37 LL tmp = fabs (X) + fabs (Y); 38 if (tmp <sum) sum = tmp; 39} 40} 41 for (int I = y/am-1; I <= y/am + 1; I ++) 42 {43 ll x = x + bm * I; 44 ll y = y-am * I; 45 if (I) 46 {47 LL tmp = fabs (X) + fabs (Y); 48 if (tmp <sum) sum = tmp; 49} 50} 51 return sum; 52} 53 54 55 int main () 56 {57 int T; 58 ll a, B, C, a, B, c, d, x, y; 59 scanf ("% d", & T); 60 while (T --) 61 {62 scanf ("% lld", & A, & B, & a, & B); 63 c = a + B; 64 C = fabs (A-B); 65 66 LL t1 = China (a, B, C ); 67 LL t2 = China (a, c, C); 68 LL t3 = China (B, c, C ); 69 70 if (t1 =-1 & t2 =-1 & t3 =-1) 71 {72 printf ("-1 \ n"); 73 continue; 74} 75 if (t1> t2) ans = t2; 76 else ans = t1; 77 78 if (ans> t3) ans = t3; 79 80 printf ("% lld \ n", ans); 81} 82 return 0; 83}View Code