Recursive sequence

Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the
cows would stand in a line, while John writes two positive numbers a
and b on a blackboard. And then, the cows would say their identity
number one by one. The first cow says the first number a and the
second says the second number b. After that, the i-th cow says the sum
of twice the (i-2)-th number, the (i-1)-th number, and i4. Now, you
need to write a program to calculate the number of the N-th cow in
order to check if John’s cows can make it right.

Input
The first line of input contains an integer t, the number of test cases. t test cases follow.
Each case contains only one line with three numbers N, a and b where N,a,b < 231 as described above.
Output
For each test case, output the number of the N-th cow. This number might be very large, so you need to output it modulo 2147493647.
Sample Input
2
3 1 2
4 1 10
Sample Output
85
369

Hint
In the first case, the third number is 85 = 21十2十3^4.
In the second case, the third number is 93 = 21十1*10十3^4 and the fourth number is 369 = 2 * 10 十 93 十 4^4.

题目大意：根据题目所给方法，求出f(n);运用矩阵快速幂。

矩阵快速幂的多种类型：f[n]=af[n-1]+bf[n-2]+c;
f[n]=af[n-1]+bf[n-2]+n^m
这题就用到了第二种类型。

Code：

#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const ll mod = 2147493647UL;//定义一个很大的变量时，编译器会报错，但使用UL之后就不会了
struct node {
	ll materix[8][8];
};

node mul(node a,node b){//矩阵乘法
	node res;
	memset(res.materix,0,sizeof(res.materix));
	for(int i=1;i<=7;i++){
		for(int j=1;j<=7;j++){
			for(int k=1;k<=7;k++){
				res.materix[i][j]=(res.materix[i][j]+a.materix[i][k]*b.materix[k][j]%mod)%mod;
			}
		}
	}
	return res;	
}

node ksm(node a,int b){//矩阵快速幂
	node ans;
	memset(ans.materix,0,sizeof(ans.materix));
	for(int i=1;i<=7;i++) ans.materix[i][i]=1;//一定要对ans进行初始化
	while(b){
		if(b&1) ans=mul(ans,a);
		a=mul(a,a);
		b>>=1;
	}
	return ans;
	
}

int main(){
	int t;
	cin>>t;
	while(t--){
		ll a;
		ll n,b;
		cin>>n>>a>>b;
		node x,y;
		x.materix[1][1]=a;x.materix[2][1]=b;x.materix[3][1]=pow(2,4);//矩阵初始化
		x.materix[4][1]=pow(2,3);x.materix[5][1]=pow(2,2);x.materix[6][1]=2,x.materix[7][1]=1;
		memset(y.materix,0,sizeof(y.materix));
		y.materix[1][2]=1;y.materix[2][1]=2;y.materix[2][2]=1;y.materix[2][3]=1;y.materix[2][4]=4;//矩阵初始化
		y.materix[2][5]=6;y.materix[2][6]=4;y.materix[2][7]=1;y.materix[3][3]=1;y.materix[3][4]=4;
		y.materix[3][5]=6;y.materix[3][6]=4;y.materix[3][7]=1;y.materix[4][4]=1;y.materix[4][5]=3;
		y.materix[4][6]=3;y.materix[4][7]=1;y.materix[5][5]=1;y.materix[5][6]=2;y.materix[5][7]=1;
		y.materix[6][6]=1;y.materix[6][7]=1;y.materix[7][7]=1;
		node ans=ksm(y,n-2);
		ans=mul(ans,x);
		cout<<ans.materix[2][1]%mod<<endl;
		
	}
	
	return 0;
}