题意：

已知\(A,B,C,D,P,n\)以及
\[ \left\{ \begin{aligned} & F_1 = A \\ & F_2 = B\\ & F_n = C*F_{n-2} + D*F_{n-2}+\lfloor(\frac{P}{n})\rfloor \end{aligned} \right. \]
，求\(F_n \ mod\ (1e9e+7)\)，\(n \leq 1e9\)

思路：

显然\(\lfloor(\frac{P}{n})\rfloor\)相同的情况是有区间的，那么直接分区间分段矩阵快速幂。

代码：

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100000 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 131;
const ll MOD = 1e9 + 7;
const double eps = 1e-8;
using namespace std;
struct Mat{ll s[3][3];Mat(){memset(s, 0, sizeof(s));}void init(){for(int i = 0; i < 3; i++)s[i][i] = 1;}
};
Mat pmul(Mat a, Mat b){Mat c;for(int i = 0; i < 3; i++){for(int j= 0; j < 3; j++){for(int k = 0; k < 3; k++){c.s[i][j] = (c.s[i][j] + a.s[i][k] * b.s[k][j]) % MOD;}}}return c;
}
Mat ppow(Mat a, ll b){Mat ret;ret.init();while(b){if(b & 1) ret = pmul(ret, a);a = pmul(a, a);b >>= 1;}return ret;
}
int main(){int T;scanf("%d", &T);while(T--){ll a, b, c, d, p, n;scanf("%lld%lld%lld%lld%lld%lld", &a, &b, &c, &d, &p, &n);Mat f, t, temp;t.s[0][0] = d, t.s[1][0] = c, t.s[2][0] = 1, t.s[0][1] = 1, t.s[2][2] = 1;f.s[0][0] = b, f.s[0][1] = a;for(int i = 3; i <= n; ){ll l, r, pn = p / i;if(pn > 0)l = i, r = min(n, (p - p % pn) / pn);elsel = i, r = n;f.s[0][2] = pn;temp = ppow(t, r - l + 1);f = pmul(f, temp);i = r + 1;}printf("%lld\n", f.s[0][0]);}return 0;
}