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 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.
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.
Source
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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 }
具体的矩阵构造会出一篇blog详细讲解
