poj 2115 简单数论题,同余方程

poj 2115 简单数论题,同余方程
/*poj 2115*/
#include
using namespace std;
long long d;
void ex_euclid(long long a,long long b,long long &x,long long &y)
{
x05if(b==0)
x05{
x05x05x=1;
x05x05y=0;
x05x05d=a;
x05x05return;
x05}
x05ex_euclid(b,a%b,x,y);
x05long long t=x;
x05x=y;
x05y=t-(a/b)*y;
}
/* c * x = b - a mod (2 ^ k) */
int main()
{
x05long long a,b,x,c,y;
x05int k;
x05while(cin>>a>>b>>c>>k)
x05{
x05x05if (a == 0 && b == 0 && c == 0 && k == 0)
break;
x05x05long long m=(1