HDU5950 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 i4i4. 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. 

InputThe 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 < 231231 as described above. 
OutputFor 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 = 2*1十2十3^4.
 In the second case, the third number is 93 = 2*1十1*10十3^4 and the fourth number is 369 = 2 * 10 十 93 十 4^4.
        
 

f(n)=f(n−1)+2∗f(n−2)+n^4
搞了一个7阶的矩阵，跑一遍快速幂就OK了。懒得画矩阵了，和这篇博客里的一样

#include <bits/stdc++.h>
#define N 7
#define mod 2147493647
using namespace std;
typedef long long ll;
struct Mat{
    ll a[N][N];
    Mat(){
        memset(a,0,sizeof(a));
        a[0][0] = 1,a[0][1] = 2,a[0][2] = 1,a[0][3] = a[0][4] = a[0][5] = a[0][6] = 0;
        a[1][0] = 1,a[1][1] = a[1][2] = a[1][3] = a[1][4] = a[1][5] = a[1][6] = 0;
        a[2][2] = 1,a[2][3] = 4,a[2][4] = 6,a[2][5] = 4,a[2][6] = 1;
        a[3][3] = 1,a[3][4] = a[3][5] = 3,a[3][6] = 1;
        a[4][4] = a[4][6] = 1,a[4][5] = 2;
        a[5][5] = a[5][6] = 1;
        a[6][6] = 1;
    }
};

Mat operator * (Mat A,Mat B){
    Mat ret;
    for (int i = 0;i < N;++i){
        for (int j = 0;j < N;++j){
            ll tmp = 0;
            for (int k = 0;k < N;++k){
                tmp = (tmp + A.a[i][k] * B.a[k][j]) % mod;
            }
            ret.a[i][j] = tmp;
        }
    }
    return ret;
}

Mat operator ^ (Mat A,int n){
    Mat ret;
    memset(ret.a,0,sizeof(ret.a));
    for (int i = 0;i < N;++i) ret.a[i][i] = 1;
    while(n){
        if (n&1) ret = ret * A;
        n >>= 1;
        A = A*A;
    }
    return ret;
}

int T,n,a,b;

int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d",&n,&a,&b);
        if (n == 1){
            printf("%d\n",a%mod);
            continue;
        }
        if (n == 2){
            printf("%d\n",b%mod);
            continue;
        }
        Mat A;
        A = A ^ (n-2);
        ll ans = (A.a[0][0] * b % mod+ A.a[0][1] * a % mod + A.a[0][2] * 81 % mod +
                    A.a[0][3] * 27 % mod + A.a[0][4] * 9 % mod + A.a[0][5] * 3 % mod + A.a[0][6] % mod) % mod;
        printf("%lld\n",ans);
    }
    return 0;
}