問題描述 　　已知遞推公式：

　　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.

　　初始值為：F(1, 1)=2, F(1, 2)=3, F(2, 1)=1, F(2, 2)=4, F(3, 1)=6, F(3, 2)=5。
　　輸入n，輸出F(n, 1)和F(n, 2)，由於答案可能很大，你只需要輸出答案除以99999999的餘數。 輸入格式 　　輸入第一行包含一個整數n。 輸出格式 　　輸出兩行，第一行為F(n, 1)除以99999999的餘數，第二行為F(n, 2)除以99999999的餘數。 樣例輸入 4 樣例輸出 14

21 資料規模和約定 　　1<=n<=10^18。

#include <stdio.h>  
#define BIG 99999999

int iskonw[4][3]={
	{0},
	{0,2,3},
	{0,1,4},
	{0,6,5}
};    

int jieguo[99999][3];//動態規劃

/*int fun(int n,int num){    //粗糙的動態規劃，題目系統裡只得了40分   
  int temp=0;
  if(n-3>=1){
	 if(jieguo[n][num]!=0) return jieguo[n][num]%BIG;
	 if(num==1)
     temp=fun(n-1,2)+2*fun(n-3,1)+5;
	 else if(num==2)
     temp=fun(n-1,1)+3*fun(n-3,1)+2*fun(n-3,2)+3;
     jieguo[n][num]=temp;
  }
  else 
	 return iskonw[n][num];
  return temp%BIG;
}       */

int fun2(long n,int num,int tag){
    long temp=0;
	long F1[4],F2[4];
	long i,j;
	F1[1]=6; F2[1]=5;
	F1[2]=1; F2[2]=4;
	F1[3]=2; F2[3]=3;

	if(n>=4)
	for(i=4;i<=n;i++){
         F1[0]=(F2[1]+2*F1[3]+5)%BIG;
		 F2[0]=(F1[1]+3*F1[3]+2*F2[3]+3)%BIG;
         for(j=3;j>=1;j--){
             F1[j]=F1[j-1];
             F2[j]=F2[j-1];
		 }
	}
    else 
		return iskonw[n][num];
	if(tag==1) return F1[0];
	return F2[0];

}

int main()  
{  
	long n=99134193;
	scanf("%d",&n);
	printf("%d\n",fun2(n,1,1));
	printf("%d\n",fun2(n,2,2));   /**/   //14 21
	return 0;
}  
 

#include<iostream>
#include<cstring>
#define MOD 99999999
using 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;
}