UESTC 288 frog appointment extension GCD

If two frogs skip step t, the coordinates of a are x + MT and B: Y + nt. To make them at the same point, the following conditions must be met: x + MT-(Y + nt) = kl (P is an integer)

Convert to: (n-m) T + KL = x-y (L> 0), then it becomes to solve the same-division equation: (n-m) T transform (x-y) mod l, solved With the extension GCD. At this time, the solution is available only when (x-y) % gcd (n-M, L) = 0.

If we have obtained a pair of x0 and Y0 for solving the homogeneous equation AX + by = m, then X0 = x * m/gcd (a, B). x0 may not be a positive integer at this time, it is not necessarily the smallest positive integer solution, so further processing is required.

Make G = gcd (A, B), for a * x0 + B * Y0 = D, there are (A/G) * x0 + (B/g) * Y0 = D/g Deformation: (A/G) (x0 + K * (B/G) + (B/g) (y0-K * (A/G) = D/g is still true. All solutions can be found based on the value of K. Therefore, X = x0 + K (B/g), so that B/g = t, x = x0 + KT, so you can use X = (x0% T + T) % t returns the minimum positive integer x.

TIPS: to prevent gcd (n-M, L) from becoming negative, first determine the positive and negative of N-M. If it is negative, then n-M is reversed to m-N, if x-y is reversed to Y-X, the correct result can still be obtained.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define Mod 1000000007#define SMod 10007#define lll __int64#define ll long longusing namespace std;#define N 1000007ll exgcd(ll a,ll b,ll &x,ll &y){    if(!b)    {        x = (ll)1,y = (ll)0;        return a;    }    ll r = exgcd(b,a%b,x,y);    ll t = x;    x = y;    y = t - a/b*y;    return r;}ll gcd(ll a,ll b){    if(!b)        return a;    return gcd(b,a%b);}int main(){    ll x,y,m,n,L;    ll kx,ky;    //freopen("1.txt","r",stdin);    //freopen("2.txt","w",stdout);    while(scanf("%lld%lld%lld%lld%lld",&x,&y,&m,&n,&L)!=EOF)    {        ll nm = n-m;        ll delta = x-y;        if(nm < 0)        {            nm = m-n;            delta = y-x;        }        ll d = exgcd(nm,L,kx,ky);        if(delta%d)        {            puts("Impossible");            continue;        }        ll t = L/d;        //printf("%lld\n",t);        ll res = (((delta/d)*kx)%t + t)%t;        printf("%lld\n",res);    }    return 0;}

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