HDu 5950 Recursive sequence（矩阵快速幂）

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1323 Accepted Submission(s): 589

Problem Description

Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers a and b on a blackboard. And then, the cows would say their identity number one by one. The first cow says the first number a and the second says the second number b. After that, the i-th cow says the sum of twice the (i-2)-th number, the (i-1)-th number, and

Input

The first line of input contains an integer t, the number of test cases. t test cases follow.
Each case contains only one line with three numbers N, a and b where N,a,b <

Output

For each test case, output the number of the N-th cow. This number might be very large, so you need to output it modulo 2147493647.

Sample Input

2 3 1 2 4 1 10

Sample Output

85 369

Hint

In the first case, the third number is 85 = 2*1十2十3^4. In the second case, the third number is 93 = 2*1十1*10十3^4 and the fourth number is 369 = 2 * 10 十 93 十 4^4.

Source

2016ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

Recommend

jiangzijing2015

题解：

　　转移方程 Fn = 2*Fn-2 + Fn-1 + n^4。如果没有这个n^4，这题就会很简单，所以我们就需要化简n^4。

　　i^4 = C(0,4) (i-1)^4 + C(1,4)*(i-1)^3 + C(2,4)*(i-1)^2 + C(3,4)*(i-1)^1 + C(4,4)*(i-1)^0

　　所以 n^4 = (n-1)^4 + 4*(n-1)^3 + 6*(n-1)^2 + 4*(i-1) + 1。

　　Fn = 2*Fn-2 + Fn-1 + (n-1)^4 + 4*(n-1)^3 + 6*(n-1)^2 + 4*(i-1) + 1。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <string>
 5 #include <algorithm>
 6 #include <cmath>
 7 #include <vector>
 8 #include <queue>
 9 #include <map>
10 #include <stack>
11 #include <set>
12 using namespace std;
13 typedef long long LL;
14 #define ms(a, b) memset(a, b, sizeof(a))
15 #define pb push_back
16 #define mp make_pair
17 const int INF = 0x7fffffff;
18 const int inf = 0x3f3f3f3f;
19 const LL mod = 2147493647;
20 const int maxn = 500+10;
21 struct Matrix
22 {
23     LL c[7][7];
24 };
25 Matrix base_Matrix =  {0,2,0,0,0,0,0,
26                        1,1,0,0,0,0,0,
27                        0,1,1,0,0,0,0,
28                        0,4,4,1,0,0,0,
29                        0,6,6,3,1,0,0,
30                        0,4,4,3,2,1,0,
31                        0,1,1,1,1,1,1};
32 Matrix mult(Matrix a, Matrix b)
33 {
34     Matrix hh = {0};
35     for(int i = 0;i<7;i++)
36         for(int j = 0;j<7;j++)
37             for(int k = 0;k<7;k++){
38                 hh.c[i][j] += (a.c[i][k]*b.c[k][j])%mod;
39                 hh.c[i][j] %= mod;
40             }
41     return hh;
42 }
43 Matrix qpow_Matrix(Matrix a, LL b)
44 {
45     Matrix base = a;
46     Matrix ans;
47     for(int i = 0;i<7;i++)
48         for(int j = 0;j<7;j++)
49             if(i==j)    ans.c[i][j] = 1;
50             else ans.c[i][j] = 0;
51 
52     while(b){
53         if(b&1) ans = mult(ans, base);
54         base = mult(base, base);
55         b>>=1;
56     }
57     return ans;
58 }
59 void init() {
60 
61 }
62 void solve() {
63     LL n, a, b;
64     cin >> n >> a >> b;
65     Matrix begin ={a,b,16,8,4,2,1};
66     if(n==1){
67         cout << a << endl;
68         return;
69     }
70     if(n==2){
71         cout << b << endl;
72         return;
73     }
74     Matrix tmp = qpow_Matrix(base_Matrix, n-2);
75     Matrix ans = mult(begin, tmp);
76 //    for(int i = 0;i<7;i++){
77 //        for(int j = 0;j<7;j++)
78 //            cout << ans.c[i][j] << " ";
79 //        cout << endl;
80 //    }
81 //    cout << endl;
82     cout << ans.c[0][1] << endl;
83 }
84 int main() {
85 #ifdef LOCAL
86     freopen("input.txt", "r", stdin);
87 //        freopen("output.txt", "w", stdout);
88 #endif
89     ios::sync_with_stdio(0);
90     cin.tie(0);
91     int T;
92     cin >> T;
93     while(T--){
94         init();
95         solve();
96     }
97     return 0;
98 }

View Code

　　具体的矩阵构造会出一篇blog详细讲解