Recursive sequence HDU - 5950

Recursive sequence

HDU - 5950

题意：求 f(n) = f(n?1)+2*f(n?2)+n4，其中 f(1)=a,f(2)=b

利用矩阵加速~

比较坑的是mod=2147493647。。。。不是2147483647，用int WA了多次=_=||

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const ll mod=2147493647;
 5 const int maxn=7;
 6
 7 struct Matrix{
 8     ll n,m[maxn][maxn];
 9
10     void init(ll sz){n=sz;memset(m,0,sizeof(m));}
11     Matrix(ll sz){init(sz);}
12     void set_I(){
13         for(int i=0;i<n;i++) m[i][i]=1;
14     }
15     Matrix operator* (Matrix a){
16         Matrix ans(n);
17         for(ll k=0;k<n;k++)
18         for(ll j=0;j<n;j++)
19         for(ll i=0;i<n;i++){
20             ans.m[i][j]=(ans.m[i][j]+m[i][k]*a.m[k][j]%mod)%mod;
21         }
22         return ans;
23     }
24 };
25
26
27 int main(){
28     int t;
29     scanf("%d",&t);
30     Matrix a(7);
31     a.set_I();
32     a.m[0][1]=2;a.m[0][2]=1;a.m[0][3]=4;
33     a.m[0][4]=6;a.m[0][5]=4;a.m[0][6]=1;
34     a.m[1][0]=1;a.m[1][1]=0;
35     a.m[2][3]=4;a.m[2][4]=6;a.m[2][5]=4;a.m[2][6]=1;
36     a.m[3][4]=3;a.m[3][5]=3;a.m[3][6]=1;
37     a.m[4][5]=2;a.m[4][6]=1;
38     a.m[5][6]=1;
39     while(t--){
40         ll n,s,b;
41         scanf("%lld%lld%lld",&n,&s,&b);
42         s%=mod;
43         b%=mod;
44         if(n==1){
45             printf("%lld\n",s);
46             continue;
47         }
48         if(n==2){
49             printf("%lld\n",b);
50             continue;
51         }
52         Matrix ans(7),p(7);
53         ans.set_I();
54         p=a;
55         n=n-2;
56         while(n){
57             if(n&1) ans=ans*p;
58             n>>=1;
59             p=p*p;
60         }
61         ll res=0;
62         res=b*ans.m[0][0]%mod;
63         res=(res+s*ans.m[0][1]%mod)%mod;
64         res=(res+ans.m[0][2]*16%mod)%mod;
65         res=(res+ans.m[0][3]*8%mod)%mod;
66         res=(res+ans.m[0][4]*4%mod)%mod;
67         res=(res+ans.m[0][5]*2%mod)%mod;
68         res=(res+ans.m[0][6])%mod;
69         printf("%lld\n",res);
70     }
71     return 0;
72 }