HDU 6395 Sequence（分段矩阵快速幂）题解

题意：

已知(A,B,C,D,P,n)以及

[left{ egin{aligned} & F_1 = A \ & F_2 = B\ & F_n = C*F_{n-2} + D*F_{n-2}+lfloor(frac{P}{n}) floor end{aligned} ight. ]

，求(F_n mod (1e9e+7))，(n leq 1e9)

思路：

显然(lfloor(frac{P}{n}) floor)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);
            else
                l = i, r = n;
            f.s[0][2] = pn;
            temp = ppow(t, r - l + 1);
            f = pmul(f, temp);
            i = r + 1;
        }

        printf("%lld
", f.s[0][0]);
    }
    return 0;
}