HDU 5950：Recursive sequence（矩阵快速幂）

http://acm.hdu.edu.cn/showproblem.php?pid=5950

题意：给出 a，b，n，递推出 f(n) = f(n-1) + f(n-2) * 2 + n ^ 4. f(1) = a, f(2) = b.

思路：在比赛时候知道是矩阵快速幂，可是推不出矩阵.那个n^4不知道怎么解决。结束后问其他人才知道要构造一个7 * 7的矩阵，而不是3 * 3的..

转自：http://blog.csdn.net/spring371327/article/details/52973534

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <string>
 #include <cmath>
 #include <queue>
 #include <vector>
 using namespace std;
 #define N 1010
 #define INF 0x3f3f3f3f
 #define MOD 2147493647
 typedef long long LL;

 struct matrix
 {
     LL a[][];

     void init() {
         memset(a, , sizeof(a));
         for(int i = ; i < ; i++) a[i][i] = ;
     }

     matrix operator * (matrix b) {
         matrix ans;
         LL tmp;
         for(int i = ; i < ; i++) {
             for(int j = ; j < ; j++) {
                 ans.a[i][j] = ;
                 for(int k = ; k < ; k++) {
                     tmp = a[i][k] * b.a[k][j] % MOD;
                     ans.a[i][j] = (ans.a[i][j] + tmp % MOD) % MOD;
                 }
             }
         }
         return ans;
     }
 };

 matrix q_pow(matrix a, LL b)
 {
     matrix ans;
     ans.init();
     while(b) {
         if(b & ) ans = ans * a;
         b >>= ;
         a = a * a;
     }
     return ans;
 }

 int main()
 {
     matrix mo;
     memset(mo.a, , sizeof(mo.a));
     mo.a[][] = ;
     mo.a[][] = ; mo.a[][] = , mo.a[][] = , mo.a[][] = , mo.a[][] = , mo.a[][] = , mo.a[][] = ;
     mo.a[][] = , mo.a[][] = , mo.a[][] = , mo.a[][] = , mo.a[][] = ;
     mo.a[][] = , mo.a[][] = , mo.a[][] = , mo.a[][] = ;
     mo.a[][] = , mo.a[][] = , mo.a[][] = ;
     mo.a[][] = , mo.a[][] = ;
     mo.a[][] = ;
     int t;
     scanf("%d", &t);
     while(t--) {
         long long n, a, b;
         scanf("%I64d%I64d%I64d", &n, &a, &b);
         if(n == ) printf("%I64d\n", a);
         else if(n == ) printf("%I64d\n", b);
         else {
             matrix ans = q_pow(mo, n - );
             LL sum = ;
             sum = (sum + ans.a[][] * a) % MOD;
             sum = (sum + ans.a[][] * b) % MOD;
             sum = (sum + ans.a[][] * ) % MOD;
             sum = (sum + ans.a[][] * ) % MOD;
             sum = (sum + ans.a[][] * ) % MOD;
             sum = (sum + ans.a[][] * ) % MOD;
             sum = (sum + ans.a[][]) % MOD;
             printf("%I64d\n", sum);
         }
     }
     return ;
 }