POJ1061(线性同余方程)

#include <iostream>
using namespace std;
typedef __int64 LL;//int前双‘_‘ 
LL extgcd(LL a,LL b,LL &x,LL &y)
{
    LL d=a;
    if(b!=0)
    {
        d=extgcd(b,a%b,y,x);
        y-=(a/b*x);
    }
    else
    {
        x=1;y=0;
    }
    return d;
}
LL gcd(LL a,LL b)
{
    if(b==0)    return a;
    else    return gcd(b,a%b);
}
LL s1,s2,v1,v2,m;
int main()
{
    while(cin>>s1>>s2>>v1>>v2>>m)
    {
        //两者相遇的条件 s1+v1*t=s2+v2*t-k*m => (v1-v2)*t+m*k=s2-s1
        //得线性同余方程 ax+by=c (a:v1-v2,x:t,b:m,k:y,c:s1-s1) 
        LL a=v1-v2;
        if(a<0)    a+=m;
        LL b=m;
        LL c=s2-s1;
        if(c<0)    c+=m;
        LL div=gcd(a,b); 
        if(c%div!=0)    //同余方程ax+by=c.有解的充要条件是 c|gcd(a,b).
        {
            cout<<"Impossible"<<endl;
            continue;
        }     
        a/=div;//将各个系数均缩小div倍
        b/=div;//ax+by=c => a‘x+b‘y=c‘ 
        c/=div; 
        LL x=0,y=0;
        extgcd(a,b,x,y);//求解线性同余方程 ax+by=1 
        x=(x*c)%b;//扩展欧几里得求的是ax+by=1中的x,结果需要将x扩大c倍 
        while(x<0)    x+=b;
        cout<<x<<endl;
    }
    return 0;
}