标签：
据说这两场加起来只要170= =而这是最简单的题目了QAQ
看到$(\frac {b + \sqrt {d} } {2} )^n$，第一反应是共轭根式$(\frac {b - \sqrt {d} } {2} )^n$
首先有$(\frac {b + \sqrt {d} } {2} )^n + (\frac {b - \sqrt {d} } {2} )^n$为整数
由高中课本知识可知，上式其实是一个三项递推数列的通项公式，而数列的递推式非常简单
$$f[x] = b * f[x - 1] - \frac{b^2 - d} {4} * f[x - 2]$$
其中$f[0] = 2, f[1] = b$，这样子直接矩阵乘法就好啦~
再看$A = (\frac {b - \sqrt {d} } {2} )^n \in [-1, 1]$
然后。。。然后发现bzoj上题面错了QAQ，实际上是"对于20%的数据$b=1,d=5$"
故$A > 0$当且仅当$d \not= b^2$且$n$为偶数，此时要将$f[n]$减掉$1$，否则不用减
1 /************************************************************** 2 Problem: 4002 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:48 ms 7 Memory:1272 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <iostream> 12 #include <cstring> 13 #include <algorithm> 14 15 using namespace std; 16 typedef unsigned long long ll; 17 const ll mod = 7528443412579576937ull; 18 19 template <class T> T sqr(T x) { 20 return x * x; 21 } 22 23 template <class T> T mul(T x, T y) { 24 static T res; 25 if (x < y) swap(x, y); 26 res = 0; 27 while (y) { 28 if (y & 1) res = (res + x) % mod; 29 x = (x << 1) % mod, y >>= 1; 30 } 31 return res; 32 } 33 34 ll b, d, n; 35 36 struct mat { 37 ll x[3][3]; 38 39 inline void clear() { 40 memset(x, 0, sizeof(x)); 41 } 42 inline void one() { 43 memset(x, 0, sizeof(x)); 44 x[1][1] = x[2][2] = 1; 45 } 46 inline void pre() { 47 this -> clear(); 48 x[1][1] = 0, x[1][2] = (d - sqr(b)) >> 2, x[2][1] = 1, x[2][2] = b; 49 } 50 51 inline ll* operator [] (int i) { 52 return x[i]; 53 } 54 55 inline mat operator * (const mat &p) const { 56 static mat res; 57 static int i, j, k; 58 for (res.clear(), i = 1; i <= 2; ++i) 59 for (j = 1; j <= 2; ++j) 60 for (k = 1; k <= 2; ++k) 61 res[i][j] = (res[i][j] + mul(x[i][k], p.x[k][j])) % mod; 62 return res; 63 } 64 inline mat& operator *= (const mat &p) { 65 return *this = *this * p; 66 } 67 } a; 68 69 inline mat pow(const mat &p, ll y) { 70 static mat x, res; 71 res.one(), x = p; 72 while (y) { 73 if (y & 1) res *= x; 74 x *= x, y >>= 1; 75 } 76 return res; 77 } 78 79 int main() { 80 cin >> b >> d >> n; 81 a.pre(); 82 a = pow(a, n); 83 cout << (((a[1][1] << 1) % mod + mul(b, a[2][1]) - (d != sqr(b) && !(n & 1))) % mod + mod) % mod << endl; 84 return 0; 85 }
标签：
原文地址：http://www.cnblogs.com/rausen/p/4448713.html
