POJ 1061 frog dating extensions Euclidean http://poj.org/problem? Id = 1061

Question: Chinese question...
Idea: it is easy to understand from the question, posx + vx * t-posy-vy * t = k * L, that is, the solution to this equation. After simplification, the equation can be written as t * (vx-vy)-k * L = posy-posx, and further reduced to k * L + t * (vy-vx) = posx-posy, L and (vx-vy) can be obtained, that is, known. This equation is a common form of Extended Euclidean equation that we are familiar.
In this case, gcdx = gcd (L, vy-vx). If (posx-posy) % gcdx = 0, the equation has a solution, that is, the frog can meet it, otherwise, it cannot be met.
Next we can use Extended Euclidean. As for the knowledge of Extended Euclidean, there are a lot of Internet resources, so I won't say much here. When we use the Extended Euclidean to obtain a group of solutions for ax + by = 1, we need the smallest positive integer solution. For the question, t must be a positive integer. If the specific solution is negative, we need to deal with it.
Ax + by = pos at this time, another group of solutions, where y is a negative number. Our goal is to change y to the smallest positive number that satisfies the equation. Now let's change the equation. Ax-kab + by + kab = pos, the equation is not changed in essence. We are simplifying it to a (x-kb) + B (y + ka) = pos, to make y positive, you only need to add several a numbers. The specific number can be calculated.
Code:
[Cpp]
# Include <iostream>
# Include <cstdio>
# Include <string. h>
Using namespace std;
 
# Define CLR (arr, val) memset (arr, val, sizeof (arr ))
Typedef long ll;
Ll xx, yy;
Ll gcd (ll a, ll B ){
If (B = 0)
Return;
Return gcd (B, a % B );
}
Void extend_Eulid (ll a, ll B ){
If (B = 0 ){
Xx = 1;
Yy = 0;
Return;
}
Else {
Extend_Eulid (B, a % B );
Int temp = xx;
Xx = yy;
Yy = temp-a/B * yy;
}
}
Int main (){
// Freopen ("1.txt"," r ", stdin );
// Freopen ("3.txt"," w ", stdout );
Ll posx, posy, vx, vy, L;
While (scanf ("% lld", & posx, & posy, & vx, & vy, & L )! = EOF ){
Ll dvx = vy-vx;
Ll dpos = posx-posy;
Ll gcdx = gcd (L, dvx );
If (dpos % gcdx ){
Printf ("Impossible \ n ");
}
Else {
Dvx = dvx/gcdx;
Dpos = dpos/gcdx;
L/= gcdx;
Extend_Eulid (L, dvx );
Yy * = dpos; www.2cto.com
Xx * = dpos;
If (L <0) L =-L;
If (yy> 0 & xx> 0) yy % = L;
Else {
Ll cnt = yy/L;
Cnt =-cnt;
Cnt ++;
Yy = (yy + L * cnt) % L;
}
Printf ("% lld \ n", yy );
}
}
Return 0;
}
Author: wmn_wmn