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; }