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//Mod linear equation system, the equation is k(m-n)(mod l)=(x-y)(mod l)
 #include<iostream>
 #include<cstdio>
 using namespace std;
 typedef long long LL;
 void ex_gcd(LL a,LL b,LL &g,LL &x,LL &y)
 {
     if(!b) {
         g=a;
         x=1;
         y=0;
     }
     else {
         ex_gcd(b,a%b,g,y,x);
         y-=(a/b)*x;
     }
 }
 int main()
 {
     std::ios::sync_with_stdio(false);
     (0);
     LL x,y,m,n,l;
     cin>>x>>y>>m>>n>>l;
     if(m==n) cout<<"Impossible"<<endl;
     else {
         if(m<n) swap(m,n), swap(x,y);
         LL g,x1,y1,cnt=y-x,ans=m-n;
         ex_gcd(ans,l,g,x1,y1);
         if(cnt%g) printf("Impossible\n");
         else {
                 LL f=cnt/g*x1,k=l/g;
                 cout<<(f%k+k)%k<<endl;
         }
     }
     return 0;
 }

Library function solves gcd:

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a,b;
    while(scanf("%d%d",&a,&b)==2)
    {
        cout<<__gcd(a,b)<<endl;
    }
    return 0;
}