POJ 2115 C Looooops （扩展欧几里德解同余方程 Ax = B(mod C) ）

题目链接：http://poj.org/problem?id=2115 题目大意：C语言循环语句，初试i赋值A，每次加C，并且模2^k，当i == B时终止，问终止循环次数，或者无法终止. 思路： 思路比较简单的一道题，解方程CX + A = B (mod 2^k)即可，变形一下：CX = B - A (mod 2^k)  

#include 
#include 
using namespace std;
long long gcd(long long a, long long b){
    return b ? gcd(b, a%b) : a;
}
void ext_gcd(long long a, long long b, long long &x, long long &y){
    if (b == 0){
        x = 1;
        y = 0;
        return ;
    }
    ext_gcd(b, a%b, x, y);
    long long tmp = x;
    x = y;
    y = tmp - a / b * y;
    return ;
}
bool equalation(long long a, long long b, long long c, long long &x, long long &y){
    long long g = gcd(a, b);
    if (c % g != 0){
        return false;
    }
    a /= g;
    b /= g;
    c /= g;
    ext_gcd(a, b, x, y);
    x *= c;
    long long tmp = abs(double(b));
    x = (x % tmp + tmp) % tmp;
    return true;
}
int main(){
    long long a, b, c, k, m;
    while(cin >> a >> b >> c >> k){
        if (a + b + c + k == 0){
            return 0;
        }
        m = 1LL << k;
        long long x, y;
        if (equalation(c, m, b-a, x, y)){
            cout << x << endl;
        }
        else{
            cout << "FOREVERn";
        }
    }
    return 0;
}