hdu 4686 矩阵乘法优化递推关系

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <set>
 5 #include <algorithm>
 6 #include <map>
 7 #include<vector>
 8 #define maxn 1010
 9 #define mod 1000000007
10 using namespace std;
11 typedef long long ll;
12 struct Matrix{
13     ll a[10][10];
14 }res,A,F,ans,temp;
15 Matrix mul(Matrix a,Matrix b){
16     memset(temp.a,0,sizeof(temp.a));
17     for(int i=0;i<5;++i){
18         for(int j=0;j<5;++j){
19             for(int k=0;k<5;++k){
20                 temp.a[i][j]+=(a.a[i][k]*b.a[k][j])%mod;
21                 temp.a[i][j]%=mod;
22              }
23         }
24     }
25     return temp;
26 }
27 void quick_pow(ll k){
28     memset(res.a,0,sizeof(res.a));
29     for(int i=0;i<5;++i)res.a[i][i]=1;
30     while(k){
31         if(k&1)res = mul(res,A);
32         A = mul(A,A);
33         k>>=1;
34     }
35 }
36 int main (){
37     ll  a0,ax,ay;
38     ll b0,bx,by;
39     ll n;
40     ll f1,a1,b1,s0;
41     while(cin>>n){
42         cin>>a0>>ax>>ay;
43         cin>>b0>>bx>>by;
44         if(n==0){printf("0\n");continue;}
45         a1 = ((a0*ax)%mod+ay)%mod;
46         b1 = ((b0*bx)%mod+by)%mod;
47         f1 = (a1*b1)%mod;
48         s0 = (a0*b0)%mod;
49         memset(A.a,0,sizeof(A.a));
50         A.a[0][0]=(ax*bx)%mod;
51         A.a[1][0]=(ax*by)%mod;
52         A.a[2][0]=(ay*bx)%mod;
53         A.a[3][0]=(ay*by)%mod;
54         A.a[1][1]=ax%mod;
55         A.a[3][1]=ay%mod;
56         A.a[2][2]=bx%mod;
57         A.a[3][2]=by%mod;
58         A.a[3][3]=1;
59         A.a[0][4]=1;
60         A.a[4][4]=1;
61         memset(F.a,0,sizeof(F.a));
62         F.a[0][0]=f1%mod;
63         F.a[0][1]=a1%mod;
64         F.a[0][2]=b1%mod;
65         F.a[0][3]=1;
66         F.a[0][4]=s0%mod;
67         quick_pow(n-1);
68         ans=mul(F,res);
69         printf("%lld\n",(ans.a[0][4])%mod);
70     }
71     return 0;
72 }