POJ 1061 The date of the frog. The minimum number of steps T for solving (X+MT)%l= (y+nt)%l.

Because is congruence, so is (X+MT)%l-(y+nt)%l=0. can be written as (x-y+ (m-n) T)%l=0. The number is a multiple of L. Then I can do this x-y+ (m-n) T + Ls = 0. Can be, s can be negative, you can meet the conditions.

#include <cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>using namespacestd;#defineINF (0X3F3F3F3F)typedefLong Long intLL; #include<iostream>#include<sstream>#include<vector>#include<Set>#include<map>#include<queue>#include<string>ll gcd (ll n,ll m) {if(n%m==0)returnm; Else returnGCD (m,n%m);} ll EXGCD (ll A,ll mod,ll&x,ll &y) {    //solving a about mod's inverse element    if(mod==0) {x=1; y=0; returnA//retains basic functionality and returns greatest common divisor} LL G=EXGCD (mod,a%mod,x,y); LL T=x;//here according to Mod==0 return back,X=y;//x, Y is the newest value x2,y2, change it, so the assignment is to X1=y2y=t-(a/mod) *y;//y1=x2 (became T)-[a/mod]y2;    returnG//retains basic functionality and returns greatest common divisor}ll get_min_number (LL a,ll b,ll c,ll&x,ll &y) {    //get the minimum integer solution of A*x+b*y=cLL ABGCD = gcd (A, b);//system GCD    if(C%ABGCD! =0)return-1;//No solutiona/= abgcd; b/= ABGCD; C/=ABGCD;    LL Tx,ty; EXGCD (A,b,tx,ty); //first get the solution of A*x+b*y=1, note this time gcd (A, b) =1x = Tx*c; y = c*ty;//any solution is because you are asking for the solution of A*x+b*y=1, but what we need is the solution of a*x+b*y=c, so at the same time multiply C//then the solution of this equation is x0=tx*c+b*k,y0=ty*c-a*k. Here's A and b about Jane. K is { -1,-2,0,1,2}, etc.LL temp_x =x; X%= b;if(x<=0) x + = b;//Minimum SolutionLL k = (temp_x-x)/b; Y+ = k*A; return 1;//1 Delegates can}voidWork () {LL x,y,n,m,l; CIN>>x>>y>>m>>n>>m; LL a=m-N; LL b=L; LL C=y-x;    LL Tx,ty; LL T=Get_min_number (a,b,c,tx,ty); if(t==-1) {cout<<"Impossible"<<Endl; return ; }    if(b<0) b=-b;  while(tx<0) TX + =b; cout<<tx<<Endl; return ;}intMain () {#ifdef local freopen ("Data.txt","R", stdin);#endifWork (); return 0;}

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★, solve the equation ax+by=c the smallest integer solution (a or b is a negative number, directly put in also no problem)

First of all, if you want to solve ax+by=c, use EXGCD can find Ax+by=1 x, Y. So we first make A and b about coprime (divided by gcd (A, A, a), we can find the Ax+by=1 x and y, but we want the solution of ax+by=c, so at the same time multiply C, where the uppercase letter is to represent about GCD (A, b) after the equation. Then the solution of this equation is x0=tx*c+b*k. Y0=ty*c-a*k. K is { -1,-2,0,1,2} and so on.

int get_min_number (int a,int b,int c,int &x,int &y)//Minimum integer solution to get A*x+b*y=c

{

int ABGCD = GCD (A, b);

if (C%ABGCD! = 0) return-1;//no solution

a/= abgcd; b/= ABGCD; C/= ABGCD;

int tx,ty;

EXGCD (A,b,tx,ty); First get the solution of A*x+b*y=1, note this time gcd (A, b) =1

x = Tx*c; y = c*ty; At the same time, C,c was a Jane.

int temp_x = x;

x%= B; if (x<=0) x + = b;//Minimum Solution

int k = (temp_x-x)/b;

Y + = k*a;

Return 1;//1 representative can

}

To get the smallest positive integer solution. Note that it is important to determine if there is a solution TLE .

int Bb=abs (B/GCD (A, b));

int Aa=abs (A/GCD (A, b));

while (x<0) x + = BB;

while (x*a+b*y!=c) y + = AA;

POJ 1061 The date of the frog. The minimum number of steps T for solving (X+MT)%l= (y+nt)%l.