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

 1 #include<bits/stdc++.h>
 2 #define F(i,a,b) for(int i=a;i<=b;i++)
 3 using namespace std;
 4 typedef long long ll;
 5
 6 const int mat_N=7;
 7 ll mo=2147493647;
 8
 9 struct mat{
10     ll c[mat_N][mat_N];
11     void init(){memset(c,0,sizeof(c));}
12     mat operator*(mat b){
13         mat M;int N=mat_N-1;M.init();
14         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;
15         return M;
16     }
17     mat operator^(ll k){
18         mat ans,M=(*this);int N=mat_N-1;ans.init();
19         F(i,0,N)ans.c[i][i]=1;
20         while(k){if(k&1)ans=ans*M;k>>=1,M=M*M;}
21         return ans;
22     }
23 }A,B,C;
24
25 int t,n,a,b;
26
27 int main()
28 {
29     scanf("%d",&t);
30     while(t--)
31     {
32         scanf("%d%d%d",&n,&a,&b);
33         A=(mat){0,1,0,0,0,0,0,
34                 2,1,1,4,6,4,1,
35                 0,0,1,4,6,4,1,
36                 0,0,0,1,3,3,1,
37                 0,0,0,0,1,2,1,
38                 0,0,0,0,0,1,1,
39                 0,0,0,0,0,0,1};
40         C=A^(n-2);
41         ll ans=0;
42         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;
43         printf("%lld\n",ans);
44     }
45     return 0;
46 }