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题目链接：hdu 4565 So Easy!
题目大意：计算题目中所给式子的值。
解题思路：矩阵快速幂。
An = (a+√b)^n, Bn = (a-√b)^n, Cn = An + Bn = (a+√b)^n+(a-√b)^n;
因为说An和Bn共轭，展开式可以互相抵消，所以保证说Cn一定是整数。
(a-1)^2 < b < a^2
-> a-1 < √b < a
-> 0 < a - √b < 1
-> 0 < (a - √b) < 1
-> [ Bn ] = 1
所以有Cn = [ An ]
Sn = Cn % m.
Sn * 2a = Sn * ((a + √b) + (a - √b)) = ((a + √b)^n + (a - √b)^n)*((a + √b) + (a - √b))
= ((a + √b)^(n+1) + (a - √b)^(n+1)) + (a^2-b)((a + √b)^(n-1) + (a - √b)^(n-1))
= Sn+1 + (a^2-b)Sn-1
Sn+1 = 2aSn + (b-a^2)Sn-1.
#include <stdio.h>
#include <string.h>
#include <math.h>#define MOD(a,b) (((a%b)+b)%b)
typedef long long ll;
const int N = 2;int m;
struct mat {ll s[N][N];mat () {memset(s, 0, sizeof(s));}void set (ll a, ll b, ll c, ll d) {s[0][0] = a; s[0][1] = b;s[1][0] = c; s[1][1] = d;}mat operator * (const mat& c) {mat ans;for (int i = 0; i < N; i++) {for (int j = 0; j < N; j++) {for (int k = 0; k < N; k++)ans.s[i][j] = MOD(ans.s[i][j] + s[i][k] * c.s[k][j], m);}}return ans;}
}A;mat powMat (int n) {mat ans;ans.set (1, 0, 0, 1);while (n) {if (n&1)ans = ans * A;n /= 2;A = A * A;}return ans;
}int main () {int a, b, n;while (scanf("%d%d%d%d", &a, &b, &n, &m) == 4) {double t = pow (a + sqrt(b), 2);ll s1 = 2 * a % m;ll s2 = (ll)ceil(t) % m;if (n <= 1) {printf("%lld\n", s1);} else {A.set (2 * a, b - a * a, 1, 0);mat ans = powMat (n-2);printf("%lld\n", MOD(s2 * ans.s[0][0] + s1 * ans.s[0][1], m));}}return 0;
}
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