poj 2115 C Looooops

这个题就是一个扩展欧几里得的运用；

B≡(A+K*C)%L 即 B = A + K*C + H*L ；B - A = KC + HL ；即线性方程为 a*x + b*y = c;

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<cstring>
#include<vector>
#define LL long long
using namespace std;
LL ExGcd( LL a, LL b , LL &x, LL &y )
{
   if( b == 0 )
   {
       x = 1;
       y = 0;
       return a;    
   }    
   LL Mod = ExGcd( b, a%b ,x ,y );
   LL temp = x;
   x = y;
   y = temp - ( a/b )*y;
   return Mod;
}
int main(  )
{
    LL A,B,C,x,y,k;
    while( scanf( "%I64d %I64d %I64d %I64d",&A,&B,&C,&k ) , A||B||C||k )
    {
        LL L = 1LL << k,p = B -A;
        LL Mod = ExGcd( C , L , x, y );
        if( p % Mod )
        {
           puts( "FOREVER" );    
        }
        else
        {
            L /= Mod ; p /= Mod;
            LL temp = ( p * x )%L;
            while( temp < 0 )
                   temp += L;
            printf( "%I64d\n",temp );    
        }
    }
    //system( "pause" );
    return 0;
}