Recursive sequence——（hdu5950）_(矩阵快速幂）

题目：点击打开链接

题意：

F(1)=a，F(2)=b，F(N)=2F(N-2)+F(N-1)+N^4

给你N,a,b(N,a,b < 2^31)

求F(N)%2147493647

分析：

如果递推式是F(N)=2F(N-2)+F(N-1)就方便了许多，基本的矩阵快速幂

但是多了n^4;

这时，又产生了i^3,i^2,i这些新项需要我们计算

由推出的这些东西 我们就可以构造矩阵了

|1   2   1   4   6   4   1|     |f(n+1) |           | f(n+2) |
|1   0   0   0   0   0   0|     |f(n)   |              |f(n+1) |
|0   0   1   4   6   4   1|     |(n+1)^4|        |(n+2)^4|
|0   0   0   1   3   3   1|  *  |(n+1)^3|   = |(n+2)^3|
|0   0   0   0   1   2   1|     |(n+1)^2|         |(n+2)^2|
|0   0   0   0   0   1   1|     | n+1   |            | n+2   |
|0   0   0   0   0   0   1|     | 1     |               |  1    |

#include <bits/stdc++.h>using namespace std;typedef long long ll;const ll mod=2147493647;int n,a,b;struct matrix{    ll m[7][7];    void init1()    {        m[0][0] = b,m[0][1] = 0,m[0][2] = 0,m[0][3] = 0,m[0][4] = 0,m[0][5] = 0,m[0][6] = 0;        m[1][0] = a,m[1][1] = 0,m[1][2] = 0,m[1][3] = 0,m[1][4] = 0,m[1][5] = 0,m[1][6] = 0;        m[2][0] = 16,m[2][1] = 0,m[2][2] = 0,m[2][3] = 0,m[2][4] = 0,m[2][5] = 0,m[2][6] = 0;        m[3][0] = 8,m[3][1] = 0,m[3][2] = 0,m[3][3] = 0,m[3][4] = 0,m[3][5] = 0,m[3][6] = 0;        m[4][0] = 4,m[4][1] = 0,m[4][2] = 0,m[4][3] = 0,m[4][4] = 0,m[4][5] = 0,m[4][6] = 0;        m[5][0] = 2,m[5][1] = 0,m[5][2] = 0,m[5][3] = 0,m[5][4] = 0,m[5][5] = 0,m[5][6] = 0;        m[6][0] = 1,m[6][1] = 0,m[6][2] = 0,m[6][3] = 0,m[6][4] = 0,m[6][5] = 0,m[6][6] = 0;    }    void init2()    {        m[0][0] = 1,m[0][1] = 2,m[0][2] = 1,m[0][3] = 4,m[0][4] = 6,m[0][5] = 4,m[0][6] = 1;        m[1][0] = 1,m[1][1] = 0,m[1][2] = 0,m[1][3] = 0,m[1][4] = 0,m[1][5] = 0,m[1][6] = 0;        m[2][0] = 0,m[2][1] = 0,m[2][2] = 1,m[2][3] = 4,m[2][4] = 6,m[2][5] = 4,m[2][6] = 1;        m[3][0] = 0,m[3][1] = 0,m[3][2] = 0,m[3][3] = 1,m[3][4] = 3,m[3][5] = 3,m[3][6] = 1;        m[4][0] = 0,m[4][1] = 0,m[4][2] = 0,m[4][3] = 0,m[4][4] = 1,m[4][5] = 2,m[4][6] = 1;        m[5][0] = 0,m[5][1] = 0,m[5][2] = 0,m[5][3] = 0,m[5][4] = 0,m[5][5] = 1,m[5][6] = 1;        m[6][0] = 0,m[6][1] = 0,m[6][2] = 0,m[6][3] = 0,m[6][4] = 0,m[6][5] = 0,m[6][6] = 1;    }};matrix mul(matrix A,matrix B){    matrix ans;    memset(ans.m,0,sizeof(ans.m));    for(int i=0; i<7; i++)    {        for(int j=0; j<7; j++)        {            for(int k=0; k<7; k++)            {                ans.m[i][j]=(ans.m[i][j]+A.m[i][k]*B.m[k][j]%mod)%mod;            }        }    }    return ans;}matrix pow(matrix A, ll p){    matrix res;    memset(res.m,0,sizeof(res.m));    for(int i=0; i<7; i++)        res.m[i][i]=1;    while(p)    {        //cout<<"___"<<endl;        if(p&1)            res=mul(res,A);        p>>=1;        A=mul(A,A);    }    return res;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&a,&b);        if(n==1)            printf("%lld\n",a%mod);        else if(n==2)            printf("%lld\n",b%mod);        else        {            matrix ans,x;            ans.init1();            x.init2();            x=pow(x,n-1);            ans=mul(x,ans);            printf("%lld\n",ans.m[1][0]);        }    }    return 0;}