蓝桥杯 递推求值

问题描述

　　已知递推公式：

　　F(n, 1)=F(n-1, 2) + 2F(n-3, 1) + 5,

　　F(n, 2)=F(n-1, 1) + 3F(n-3, 1) + 2F(n-3, 2) + 3.

　　初始值为：F(1, 1)=2, F(1, 2)=3, F(2, 1)=1, F(2, 2)=4, F(3, 1)=6, F(3, 2)=5。
　　输入n，输出F(n, 1)和F(n, 2)，由于答案可能很大，你只需要输出答案除以99999999的余数。

输入格式

　　输入第一行包含一个整数n。

输出格式

　　输出两行，第一行为F(n, 1)除以99999999的余数，第二行为F(n, 2)除以99999999的余数。

样例输入

4

样例输出

14

21

数据规模和约定

　　1<=n<=10^18。

这个题数据量很大，如果直接递推的话，一定会超时，所以采用矩阵快速幂的方法，将时间复杂度由o(n)降低为o(logn),不懂矩阵快速幂的可自行百度。

#include<iostream>#include<cstring>#define MOD 99999999using namespace std;typedef  long long**  Mat;void mul_mat(Mat a, Mat b, Mat &c, int R_number_a, int C_number_a, int C_number_b){int i, j, k;Mat C = new long long*[R_number_a];for (int i = 0;i<R_number_a;i++){C[i] = new long long[C_number_b];for (int j = 0;j<C_number_b;j++)C[i][j] = 0ll;}for ( i = 0;i<R_number_a;i++)for ( k = 0;k < C_number_a;k++)for ( j = 0;j<C_number_b;j++)C[i][j] += a[i][k] * b[k][j]%MOD;delete c;c = C;}Mat pow_mat(Mat region, int R_number, int C_number, long long n){Mat ans = new long long*[R_number];Mat t = new long long*[R_number];for (int i = 0;i<R_number;i++){ans[i] = new long long[C_number];for (int j = 0;j<C_number;j++)ans[i][j] = (long long)(i == j);}for (int i = 0;i<R_number;i++){t[i] = new long long[C_number];for (int j = 0;j<C_number;j++)t[i][j] = region[i][j];}while (n){if (n & 1)mul_mat(ans, t, ans, R_number, C_number, C_number);mul_mat(t, t, t, R_number, C_number, C_number);n = n >> 1;}return ans;}int main(){Mat region = new long long*[7], ans;long long n, a[7] = {5ll,6ll,4ll,1ll,3ll,2ll,1ll};cin >> n;region[0] = new long long[7];region[1] = new long long[7];for (int i = 0;i < 7;i++)region[i] = new long long[7],memset(region[i],0,sizeof(long long)*7);region[0][1]= region[1][0]= region[2][0]= region[3][1]= region[4][2]= region[5][3]= region[6][6]=1ll;region[0][4] = 2ll;region[0][5] = 3ll;region[0][6] = 3ll;region[1][5] = 2ll;region[1][6] = 5ll;if (n < 4) {switch (n) {case 1:cout << 2 << endl << 3 << endl;break;case 2:cout << 1 << endl << 4<< endl;break;case 3:cout << 6 << endl << 5 << endl;break;}}else {ans = pow_mat(region, 7, 7, n - 3);long long fn1 = 0ll, fn2 = 0ll;for (int i = 0;i < 7;i++)fn1 += ans[1][i] * a[i], fn2+= ans[0][i] * a[i];cout << fn1%MOD << endl << fn2%MOD << endl;}return 0;}