POJ 2115 C Looooops

被虐得很惨的一道题，WA了无数次，根据题意推出 C * x + (2 ^ k) * y = B - A，用拓展欧几里德求出x的最小值。如果无解则是死循环！

 

/*Accepted    164K    0MS    C++    882B    2012-07-19 12:03:12*/



#include<cstdio>

#include<cstring>

#include<cstdlib>

#define LL long long





LL extgcd( LL a, LL b, LL &x, LL &y)

{

    if( b == 0) { x = 1; y = 0; return a;}

    LL d = extgcd( b, a % b, x, y);

    LL t = x;

    x = y;

    y = t - a / b * y;

    return d;

}



int main()

{

    LL a, b, c, e, n;

    int k;

    LL x, y;

    while( scanf("%lld%lld%lld%d", &a, &b, &c, &k) == 4)

    {

        if( a == 0 && b == 0 && c == 0 && k == 0)

            break;

        n = 1LL << k ;

        e = extgcd( c, n, x, y);

        LL t = b - a ;

        if( t % e != 0)

        {

            printf( "FOREVER\n");

        }

        else{

            x = t / e * x; // x = ((t / e) * x) % n + n

            x = ( x % (n / e) + (n / e)) % (n / e); //x = x % (n / e);

            printf( "%lld\n", x) ;

        }

    }

    return 0;

}