题目链接：

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;}