标签:ext ret strong eve iostream pre end style cout
思路:
求最小的非负整数t使得C * t % 2k = (B - A) % 2k。
实现:
1 #include <iostream> 2 #include <cstdio> 3 4 using namespace std; 5 typedef long long ll; 6 7 ll abs(ll x) 8 { 9 return x < 0 ? -x : x; 10 } 11 12 ll extgcd(ll a, ll b, ll & x, ll & y) 13 { 14 int d = a; 15 if (!b) 16 { 17 x = 1; y = 0; 18 } 19 else 20 { 21 d = extgcd(b, a % b, y, x); 22 y -= (a / b) * x; 23 } 24 return d; 25 } 26 27 int main() 28 { 29 ll a, b, c, k; 30 while (cin >> a >> b >> c >> k, a || b || c || k) 31 { 32 ll tmp = 1; 33 tmp <<= k; 34 ll d, x, y; 35 d = extgcd(c, tmp, x, y); 36 if ((b - a) % d) puts("FOREVER"); 37 else 38 { 39 ll mod = abs(tmp / d); 40 x = (x * (b - a) / d % mod + mod) % mod; 41 cout << x << endl; 42 } 43 } 44 return 0; 45 }
标签:ext ret strong eve iostream pre end style cout
原文地址:http://www.cnblogs.com/wangyiming/p/7188450.html