hdu_5950_Recursive sequence(矩阵快速幂)

题目链接：hdu_5950_Recursive sequence

题意：递推求解：F(n) = 2*F(n-2) + F(n-1) + n⁴ 和F(1) = a，F(2) = b；

题解：

一看数据范围，肯定矩阵加速递推，不过公式不是线性的，需要把公式转换为线性的公式

#include<bits/stdc++.h>
#define F(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
typedef long long ll;

const int mat_N=7;
ll mo=2147493647;

struct mat{
    ll c[mat_N][mat_N];
    void init(){memset(c,0,sizeof(c));}
    mat operator*(mat b){
        mat M;int N=mat_N-1;M.init();
        F(i,0,N)F(j,0,N)F(k,0,N)M.c[i][j]=(M.c[i][j]+c[i][k]*b.c[k][j])%mo;
        return M;
    }
    mat operator^(ll k){
        mat ans,M=(*this);int N=mat_N-1;ans.init();
        F(i,0,N)ans.c[i][i]=1;
        while(k){if(k&1)ans=ans*M;k>>=1,M=M*M;}
        return ans;
    }
}A,B,C;

int t,n,a,b;

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&a,&b);
        A=(mat){0,1,0,0,0,0,0,
                2,1,1,4,6,4,1,
                0,0,1,4,6,4,1,
                0,0,0,1,3,3,1,
                0,0,0,0,1,2,1,
                0,0,0,0,0,1,1,
                0,0,0,0,0,0,1};
        C=A^(n-2);
        ll ans=0;
        ans=(C.c[1][0]*a+C.c[1][1]*b+C.c[1][2]*16+C.c[1][3]*8+C.c[1][4]*4+C.c[1][5]*2+C.c[1][6])%mo;
        printf("%lld
",ans);
    }
    return 0;
}