题目链接：https://vjudge.net/problem/HDU-5950
Recursive sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2727 Accepted Submission(s): 1226
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
23 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）
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jiangzijing2015
题意：
求 f(n) = f(n−1) + 2*f(n−2) + n^4，其中 f(1)=a,f(2)=b
题解：
典型的矩阵快速幂的运用。关键是i^4怎么维护？我们可以当成求第i+1项，那么i^4就变成了（i+1）^4。那么这时我们可以用二项式定理从i^4、i^3、i^2、i^1、i^0的组合中得到（i+1）^4。也就是说总共需要维护：f[i+1]、f[i]、（i+1）^4、（i+1）^3、（i+1）^2、（i+1）^1、（i+1）^0。矩阵如下：
代码如下:
1 #include <bits/stdc++.h>
2 #define rep(i,s,t) for(int (i)=(s); (i)<=(t); (i)++)
3 #define ms(a,b) memset((a),(b),sizeof((a)))
4 using namespace std;
5 typedef long long LL;
6 const LL mod = 2147493647;
7 const int maxn = 1e5;
8
9 struct Mat
10 {
11 LL mat[7][7];
12 void init()
13 {
14 rep(i,0,6) rep(j,0,6)
15 mat[i][j] = (i==j);
16 }
17 };
18
19 Mat p = { 1, 2, 1, 4, 6, 4, 1,
20 1, 0, 0, 0, 0, 0, 0,
21 0, 0, 1, 4, 6, 4, 1,
22 0, 0, 0, 1, 3, 3, 1,
23 0, 0, 0, 0, 1, 2, 1,
24 0, 0, 0, 0, 0, 1, 1,
25 0, 0, 0, 0 ,0, 0, 1
26 };
27
28 Mat mul(Mat x, Mat y)
29 {
30 Mat s;
31 ms(s.mat,0);
32 rep(i,0,6) rep(j,0,6) rep(k,0,6)
33 s.mat[i][j] += (x.mat[i][k]*y.mat[k][j])%mod, s.mat[i][j] %= mod;
34 return s;
35 }
36
37 Mat qpow(Mat x, LL y)
38 {
39 Mat s;
40 s.init();
41 while(y)
42 {
43 if(y&1) s = mul(s, x);
44 x = mul(x, x);
45 y >>= 1;
46 }
47 return s;
48 }
49 int main()
50 {
51 int T;
52 scanf("%d",&T);
53 while(T--)
54 {
55 LL n, a, b;
56 scanf("%lld%lld%lld",&n,&a,&b);
57 if(n == 1)
58 {
59 printf("%lld\n",a);
60 continue;
61 }
62 if(n == 2)
63 {
64 printf("%lld\n",b);
65 continue;
66 }
67
68 Mat x = p;
69 x = qpow(x, n-2);
70
71 LL ans = 0;
72 ans = (ans + b*x.mat[0][0]) % mod;
73 ans = (ans + a*x.mat[0][1]%mod) % mod;
74 ans = (ans + 16*x.mat[0][2]%mod) % mod;
75 ans = (ans + 8*x.mat[0][3]%mod) % mod;
76 ans = (ans + 4*x.mat[0][4]%mod) % mod;
77 ans = (ans + 2*x.mat[0][5]%mod) % mod;
78 ans = (ans+x.mat[0][6]) % mod;
79 printf("%lld\n",ans);
80 }
81 }
View Code
