hdu2802：F(N)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2802

参考思路：通过变换，有F(N+2) = F(N) + 3 * N * N + 9 * N + 7，于是可以构造递推矩阵乘法：

源代码：

#include <iostream> #include <algorithm> #include <cstdio> using namespace std; int matrix1[4][4] = {1, 1, 1, 1}; int matrix2[4][4] = {7, 4, 2, 1}; int result[4][4]; void copyMatrix(int lhs[][4], const int rhs[][4]) { for(int i=0; i!=4; i++) { for(int j=0; j!=4; j++) lhs[i][j] = rhs[i][j]; } } void mul(int lhs[][4], int rhs[][4]) { int temp[4][4]; int i, j, k; for(i=0; i!=4; i++) { for(j=0; j!=4; j++) { temp[i][j] = 0; for(k=0; k!=4; k++) { temp[i][j] = (temp[i][j] + lhs[i][k]*rhs[k][j])%2009; } } } copyMatrix(lhs, temp); } int main() { int n; while(scanf("%d", &n) && n!=0) { int k; int help[4][4] = {1, 0, 0, 0, 3, 1, 0, 0, 9, 4, 1, 0, 7, 4, 2, 1}; if(n%2 == 1) { copyMatrix(result, matrix1); k = (n-1)/2; } else { copyMatrix(result, matrix2); k = (n-2)/2; } while(k != 0) { if((k&1)==1) { mul(result, help); } mul(help, help); k >>= 1; } printf("%d\n", result[0][0]%2009); } return 0; } </cstdio></algorithm></iostream>