题目链接：传送门
题目：
Recursive sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 4173 Accepted Submission(s): 1813 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 i4. Now, you need to write a program to calculate the number of the N-th cow in order to check if John’s cows can make it right. 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 < 231 as described above. 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.
题目大意：
已知a1 = a，a2 = b，ai = ai-1 + 2ai-2 + n4，求aN，答案对2147493647取模。
N，a，b ＜ 231。
思路：
因为N超级大，考虑快速幂，但是通项怎么都搞不出来。
又题目给的是递推式，考虑用矩阵快速幂。。。。
令Fn = [an-1, an]，则Fn-1 = [an-2，an-1]，但是这样an+1就算不上n4了。那就把n4加上去试试看：
再令Fn = [an-1，an，(n+1)4]，这样就可以推出an+1了，但是(n+1)4又不能递推。。。展开(n+1)4发现：(n+1)4 = n4 + 4n3 + 6n2 + 4n + 1，可以由n4、n3、n2、n、1递推出来。同时(n+1)3、(n+1)2、n+1、1都可以用n4、n3、n2、n、1递推出来，所以Fn和系数矩阵base就出来了：
Fn = [an-1，an，(n+1)4，(n+1)3，(n+1)2，n+1，1]；
$\begin{bmatrix}
0 &2 &0 &0 &0 &0 &0 \\
1 &1 &0 &0 &0 &0 &0 \\
0 &1 &1 &0 &0 &0 &0 \\
0 &0 &4 &1 &0 &0 &0 \\
0 &0 &6 &3 &1 &0 &0 \\
0 &0 &4 &3 &2 &1 &0 \\
0 &0 &1 &1 &1 &1 &1
\end{bmatrix}$
代码：
#include <bits/stdc++.h> using namespace std; typedef long long ll; #define MAXN 7 const ll mod = 2147493647; struct Matrix { ll mat[MAXN][MAXN]; Matrix() {} Matrix operator*(Matrix const &b)const { Matrix res; memset(res.mat, 0, sizeof(res.mat)); for (int i = 0 ;i < MAXN; i++) for (int j = 0; j < MAXN; j++) for (int k = 0; k < MAXN; k++) res.mat[i][j] = (res.mat[i][j]+this->mat[i][k] * b.mat[k][j])%mod; return res; } }base; Matrix pow_mod(Matrix base, ll n) { Matrix res; memset(res.mat, 0, sizeof(res.mat)); for (int i = 0; i < MAXN; i++) res.mat[i][i] = 1; while (n > 0) { if (n & 1) res = res*base; base = base*base; n >>= 1; } return res; }//输入基础矩阵返回n次幂的矩阵; //res.mat[0][0] 就是最终的值F(N+1) //注意修改MAXN和mod void init() { base.mat[0][0] = 0; base.mat[0][1] = 2; base.mat[0][2] = 0; base.mat[0][3] = 0; base.mat[0][4] = 0; base.mat[0][5] = 0; base.mat[0][6] = 0; base.mat[1][0] = 1; base.mat[1][1] = 1; base.mat[1][2] = 0; base.mat[1][3] = 0; base.mat[1][4] = 0; base.mat[1][5] = 0; base.mat[1][6] = 0; base.mat[2][0] = 0; base.mat[2][1] = 1; base.mat[2][2] = 1; base.mat[2][3] = 0; base.mat[2][4] = 0; base.mat[2][5] = 0; base.mat[2][6] = 0; base.mat[3][0] = 0; base.mat[3][1] = 0; base.mat[3][2] = 4; base.mat[3][3] = 1; base.mat[3][4] = 0; base.mat[3][5] = 0; base.mat[3][6] = 0; base.mat[4][0] = 0; base.mat[4][1] = 0; base.mat[4][2] = 6; base.mat[4][3] = 3; base.mat[4][4] = 1; base.mat[4][5] = 0; base.mat[4][6] = 0; base.mat[5][0] = 0; base.mat[5][1] = 0; base.mat[5][2] = 4; base.mat[5][3] = 3; base.mat[5][4] = 2; base.mat[5][5] = 1; base.mat[5][6] = 0; base.mat[6][0] = 0; base.mat[6][1] = 0; base.mat[6][2] = 1; base.mat[6][3] = 1; base.mat[6][4] = 1; base.mat[6][5] = 1; base.mat[6][6] = 1; } int main() { init(); int T; cin >> T; while (T--) { ll n, a, b; scanf("%lld%lld%lld", &n, &a, &b); Matrix F2; memset(F2.mat, 0, sizeof F2.mat); F2.mat[0][0] = a; F2.mat[0][1] = b; F2.mat[0][2] = 3*3*3*3; F2.mat[0][3] = 3*3*3; F2.mat[0][4] = 3*3; F2.mat[0][5] = 3; F2.mat[0][6] = 1; Matrix FN = F2*pow_mod(base, n-2); cout << FN.mat[0][1] << endl; } return 0; }