POJ 1061 Frog Dating (extended Euclid)

Main topic:

That is, give you two starting x and Y, then x increments m,y each time n, as well as length L, to find the smallest number of changes T, there is (x+m*t)-(y+n*t) ==p*l.

Problem Solving Ideas:

Naked naked expansion of Euclid.

Analysis: Assume that after jumping the T-time, the Frog 1 coordinates is x+m*t, Frog 2 coordinates of Y+N*T. They can meet in the case of (x+m*t)-(y+n*t) ==p*l, where P is an integer, deform a bit

Get (N-M) *t+p*l==x-y we set a= (n-m), b=l,c=x-y,t=x,p=y. So we get ax+by==c. Excited ah, this is not the same formula above.

Directly apply the extended Euclidean function to get a set of solutions for X, Y. Since the question is how many times to jump at least, then only X is the information we need. So, do you think X is the smallest?

The answer is probably not!  So how do we get the smallest solution? We consider the formula of all the solutions of x: X=x0+b/d*t. X0 is what we just asked for, obviously there is a monotone function on the right, and when T is a number opposite to x positive or negative, the smallest x can be obtained. So that the positive and negative properties of X are positive, then X=X0-B/D*T1 (T1==-T). Make x==0, then t=x0*d/b, the smallest x equals x0 minus t*b/d. Here the X may be negative, if it is negative, we add a b/d to it is the answer!

Code:

1# include<cstdio>2# include<iostream>3# include<fstream>4# include<algorithm>5# include<functional>6# include<cstring>7# include<string>8# include<cstdlib>9# include<iomanip>Ten# include<numeric> One# include<cctype> A# include<cmath> -# include<ctime> -# include<queue> the# include<stack> -# include<list> -# include<Set> -# include<map> +  - using namespacestd; +  A Const DoublePi=4.0*atan (1.0); at  -typedefLong LongLL; -typedef unsignedLong LongULL; -  -# define INF999999999 -  in LL x,y,a,b,c,d; - LL n,m,x,y,l; to  + ll EXT_GCD (ll A,ll b) - { the LL t,d; *     if(b==0 ) $     {Panax Notoginsengx =1; -y =0; the         returnA; +     } AD = EXT_GCD (b,a%b); thet =x; +x =y; -y = t (A/b) *y; $     returnD; $ } -  -  the intMainvoid) - {Wuyi      while(cin>>x>>y>>m>>n>>L) the     { -A = nm; Wub =L; -c = XY; AboutD =EXT_GCD (A, b); $         if(c%d!=0 ) -         { -printf"impossible\n"); -             Continue; A  +         } thex = x* (c/d); -y = y* (c/d); $  theLL k = x*d/b; theK = x-k*b/D; the         if(k<0 ) the         { -k+=b/D; in         } thecout<<k<<Endl; the     } About  the  the     return 0; the}

POJ 1061 Frog Dating (extended Euclid)