HDU 6395 Sequence 【矩阵快速幂 && 暴力】

任意门：http://acm.hdu.edu.cn/showproblem.php?pid=6395

Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2564    Accepted Submission(s): 999

Problem Description

Let us define a sequence as below

⎧⎩⎨⎪⎪⎪⎪⎪⎪F1F2Fn===ABC⋅Fn−2+D⋅Fn−1+⌊Pn⌋

  Your job is simple, for each task, you should output Fn module 109+7.

 

Input

The first line has only one integer T, indicates the number of tasks.

Then, for the next T lines, each line consists of 6 integers, A , B, C, D, P, n.

1≤T≤200≤A,B,C,D≤1091≤P,n≤109

 

Sample Input

2

3 3 2 1 3 5

3 2 2 2 1 4

 

Sample Output

36

24

 

Source

2018 Multi-University Training Contest 7

 

题意概括：

给出 A，B，C，D，P，N；

根据函数：

F(1)=A， F(2)=B,  F(i)=C*F(i-2)+D*F(i-1)+p/i；

求 F( N );

解题思路：

一开始看错题目，以为 p/n 为 一个常数，其实题目里的 n 是变量（即题意里的 i ）；

如果是常数直接构造矩阵，矩阵快速幂跑一波即可，但是这里是是变量。

所以一开始选择了暴力 p/i ；p的范围是 1e9 果断超时。

怎么优化呢？

其实由于整型除法的向下取整，我们可以按 p/i 的种类分成一段一段的，这样大大缩短了暴力区间。

AC code：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#define LL long long
using namespace std;
const int MAXN = 11;
const int Mod = 1e9+7;
const int NN = 3;
int N, A, B, C, D, P;
struct mat
{
    LL m[MAXN][MAXN];
}base, ans;

mat muti(mat a, mat b)
{
    mat res;
    memset(res.m, 0, sizeof(res.m));
    for(int i = 1; i <= NN; i++)
    for(int j = 1; j <= NN; j++){
            if(a.m[i][j]){
                for(int k = 1; k <= NN; k++){
                    res.m[i][k] = (res.m[i][k] + a.m[i][j]*b.m[j][k])%Mod;
                }
            }
        }

    return res;
}

mat qpow(mat a, int n)
{
    mat res;
    memset(res.m, 0, sizeof(res.m));
    for(int i = 1; i <= NN; i++) res.m[i][i] = 1;
    while(n){
        if(n&1) res = muti(res, a);
        n>>=1;
        a = muti(a, a);
    }
    return res;
}


int main()
{
    int K, T_case;
    scanf("%d", &T_case);
    while(T_case--){
        memset(base.m, 0, sizeof(base.m));
        memset(ans.m, 0, sizeof(ans.m));
        scanf("%d %d %d %d %d %d", &A, &B, &C, &D, &P, &N);
        if(N == 1){printf("%d
", A);continue;}
        if(N == 2){printf("%d
", B);continue;}
        else{
            base.m[1][2] = C;
            base.m[2][2] = D;
            base.m[2][1] = 1;
            base.m[3][3] = 1;
            base.m[3][2] = P/3;
            ans.m[1][1] = A;
            ans.m[1][2] = B;
            ans.m[1][3] = 1;
            int now = 3, x, len = 0, lst;
            for(;now <= N; now = lst+1){
                x = P/now;
                if(x != 0) lst = min(P/x, N);
                else lst = N;
                len = lst-now+1;
                base.m[3][2] = x;
                ans = muti(ans, qpow(base, len));
            }
        }
        printf("%lld
", ans.m[1][2]);
    }

    return 0;
}