Solve ax + by = C equations like this

Basic knowledge:

1. For any ax+by=c, if we know there is a group of solutions x0, y0; Then x1 = x0+kb ' (b ' =b/gcd (A, a)), y1 = Y0-ka ' (a ' =a/gcd (b));

The process for solving ax + by = C is as follows:

1. First we use EGCD to find the solution of ax+by=g (g = gcd (A, b)). Using this algorithm we can find three numbers g, X, Y

2. Then we judge C%g==0? If it is not equal to 0, then the equation has no integer solution. If it equals 0, then take the third step.

3. Using g, X, Y, c we find a group of ax+by=c solution x0 = x*c/g, y0 = y*c/g;

4. We now use the basic knowledge 1 to solve the ax+by=c of any set of solutions. We can take advantage of modulo operations when we are seeking the smallest possible x of satisfying conditions:

Example: POJ1061

Test instructions is there are two frogs jumping on the equator, the first frog's starting point is X, each jump m, the second starting from Y each jump n meters, the equator length is L, asked two frogs at least a few steps meet?

We can list the following equations x+km = y+kn mod (l) = + K (m-n) + k1*l = y-x, the code is as follows:

#include <cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespaceStd;typedefLong LongLL;voidGCD (ll A, ll B, LL &d, LL &x, LL &y) {    if(!B) {d=a; x=1; y=0;} Else{gcd (b, a%b, D, y, x); y-= x* (A/b); }}intMain () {LL x, y, M, N, L; CIN>>x>>y>>m>>n>>l; LL a=m-n, B=l, c=y-x;    LL G;    GCD (A, B, G, x, y); if(C%g! =0) {cout<<"Impossible"<<Endl; return 0; } LL x0= c/g*x; //x1 = x0+k*b/gLL BG = b/g>0? b/g:-b/G; X0= (X0%BG+BG)%BG; Here we use the modulo operation to solve the minimum value cout<<x0<<Endl; return 0;}

Solve ax + by = C equations like this