HDU5950（矩阵快速幂）
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5950
题意：f(n) = f(n-1) + 2*f(n-2) + n^4，f(1) = a , f(2) = b，求f(n)
思路：对矩阵快速幂的了解仅仅停留在fib上，重现赛自己随便乱推还一直算错，快两个小时才a还wa了好几次....
主要就是构造矩阵：(n+1)^4 = n^4 + 4n^3 + 6n^2 + 4n + 1
|1 2 1 4 6 4 1| | f(n+1) | | f(n+2) |
|1 0 0 0 0 0 0| | f(n) | | f(n+1) |
|0 0 1 4 6 4 1| | (n+1)^4 | | (n+2)^4 |
|0 0 0 1 3 3 1| * | (n+1)^3 | = | (n+2)^3 |
|0 0 0 0 1 2 1| | (n+1)^2 | | (n+2)^2 |
|0 0 0 0 0 1 1| | n+1 | | n+2 |
|0 0 0 0 0 0 1| | 1 | | 1 |
1 #include<cstdio> 2 using namespace std; 3 typedef long long ll; 4 const ll mod = 2147493647; 5 ll n,a,b; 6 struct Matrix 7 { 8 ll m[7][7]; 9 void init1() 10 { 11 m[0][0] = b,m[0][1] = 0,m[0][2] = 0,m[0][3] = 0,m[0][4] = 0,m[0][5] = 0,m[0][6] = 0; 12 m[1][0] = a,m[1][1] = 0,m[1][2] = 0,m[1][3] = 0,m[1][4] = 0,m[1][5] = 0,m[1][6] = 0; 13 m[2][0] = 16,m[2][1] = 0,m[2][2] = 0,m[2][3] = 0,m[2][4] = 0,m[2][5] = 0,m[2][6] = 0; 14 m[3][0] = 8,m[3][1] = 0,m[3][2] = 0,m[3][3] = 0,m[3][4] = 0,m[3][5] = 0,m[3][6] = 0; 15 m[4][0] = 4,m[4][1] = 0,m[4][2] = 0,m[4][3] = 0,m[4][4] = 0,m[4][5] = 0,m[4][6] = 0; 16 m[5][0] = 2,m[5][1] = 0,m[5][2] = 0,m[5][3] = 0,m[5][4] = 0,m[5][5] = 0,m[5][6] = 0; 17 m[6][0] = 1,m[6][1] = 0,m[6][2] = 0,m[6][3] = 0,m[6][4] = 0,m[6][5] = 0,m[6][6] = 0; 18 } 19 void init2() 20 { 21 m[0][0] = 1,m[0][1] = 2,m[0][2] = 1,m[0][3] = 4,m[0][4] = 6,m[0][5] = 4,m[0][6] = 1; 22 m[1][0] = 1,m[1][1] = 0,m[1][2] = 0,m[1][3] = 0,m[1][4] = 0,m[1][5] = 0,m[1][6] = 0; 23 m[2][0] = 0,m[2][1] = 0,m[2][2] = 1,m[2][3] = 4,m[2][4] = 6,m[2][5] = 4,m[2][6] = 1; 24 m[3][0] = 0,m[3][1] = 0,m[3][2] = 0,m[3][3] = 1,m[3][4] = 3,m[3][5] = 3,m[3][6] = 1; 25 m[4][0] = 0,m[4][1] = 0,m[4][2] = 0,m[4][3] = 0,m[4][4] = 1,m[4][5] = 2,m[4][6] = 1; 26 m[5][0] = 0,m[5][1] = 0,m[5][2] = 0,m[5][3] = 0,m[5][4] = 0,m[5][5] = 1,m[5][6] = 1; 27 m[6][0] = 0,m[6][1] = 0,m[6][2] = 0,m[6][3] = 0,m[6][4] = 0,m[6][5] = 0,m[6][6] = 1; 28 } 29 Matrix operator * (Matrix t) 30 { 31 Matrix res; 32 for (int i = 0; i < 7; i++) 33 { 34 for (int j = 0; j < 7; j++) 35 { 36 res.m[i][j] = 0; 37 for (int k = 0;k < 7; k++) 38 res.m[i][j] = (res.m[i][j] + (m[i][k] % mod) * (t.m[k][j] % mod) % mod) % mod; 39 } 40 } 41 return res; 42 } 43 Matrix operator ^ (int k) 44 { 45 Matrix res,s; 46 res.init2(); 47 s.init2(); 48 while(k) 49 { 50 if(k & 1) 51 res = res * s; 52 k >>= 1; 53 s = s * s; 54 } 55 return res; 56 } 57 }; 58 int main() 59 { 60 int T; 61 scanf("%d",&T); 62 while(T--) 63 { 64 scanf("%lld %lld %lld",&n,&a,&b); 65 if(n == 1) 66 { 67 printf("%lld\n",a % mod); 68 continue; 69 } 70 if(n == 2) 71 { 72 printf("%lld\n",b % mod); 73 continue; 74 } 75 Matrix ans,t; 76 ans.init1(); 77 t.init2(); 78 ans = (t^(n-3)) * ans; 79 printf("%lld\n",ans.m[0][0]); 80 } 81 return 0; 82 }
