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 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 = 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.

学习用矩阵快速幂做这种递推问题构造矩阵是很重要的还是要多练

给出递推公式 fn=fn−1+2fn−2+n4f1=a,f2=b

要构造一个矩阵A使得

$> [> > f n f n - 1 n 4 n 3 n 2 n 1 > >] > = > [> > f n - 1 f n - 2 (n - 1) 4 (n - 1) 3 (n - 1) 2 (n - 1) 1 > >] > \times > A >$
根据矩阵乘法可以得到
$> A = > ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ > > 1 > 2 > 1 > 4 > 6 > 4 > 1 > 100000000146410001331000012100000110000001 > ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ >$

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
typedef long long ll;
using namespace std;
const int N=7;
const ll MOD=2147493647;
struct Mat{
    ll m[N][N];
};
Mat operator * (Mat a,Mat b){
    Mat t;
    memset(t.m,0,sizeof(t.m));
    for(int k=0;k<N;k++){
        for(int i=0;i<N;i++){
            if(a.m[i][k]==0){
                continue;
            }
            for(int j=0;j<N;j++){
                if(b.m[k][j]==0){
                    continue;
                }
                t.m[i][j]=(t.m[i][j]+(a.m[i][k]%MOD)*(b.m[k][j]%MOD))%MOD;
            }
        }
    }
    return t;
}
Mat operator ^ (Mat a,int n){
    Mat t;
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            t.m[i][j]=i==j?1:0;
        }
    }
    while(n){
        if(n%2){
            t=t*a;
        }
        n/=2;
        a=a*a;
    }
    return t;
}
int main(void){
    int t;
    scanf("%d",&t);
    ll n,a,b;
    Mat k;
    memset(k.m,0,sizeof(k.m));
    k.m[0][0]=k.m[0][1]=k.m[2][0]=k.m[2][2]=k.m[3][3]=k.m[4][4]=k.m[5][5]=k.m[6][0]=k.m[6][2]=k.m[6][3]=k.m[6][4]=k.m[6][5]=k.m[6][6]=1;
    k.m[1][0]=k.m[5][4]=2;
    k.m[4][3]=k.m[5][3]=3;
    k.m[3][0]=k.m[3][2]=k.m[5][0]=k.m[5][2]=4;
    k.m[4][0]=k.m[4][2]=6;
    Mat g;
    while(t--){
        scanf("%lld%lld%lld",&n,&a,&b);
        if(n==1){
            printf("%lld\n",a%MOD);
        }
        else if(n==2){
            printf("%lld\n",b%MOD);
        }
        else{
            memset(g.m,0,sizeof(g));
            g.m[0][0]=b;
            g.m[0][1]=a;
            g.m[0][2]=16;
            g.m[0][3]=8;
            g.m[0][4]=4;
            g.m[0][5]=2;
            g.m[0][6]=1;
            printf("%lld\n",(g*(k^(n-2))).m[0][0]%MOD);
        }
    }
    return 0;
}