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;
}