线性递推式（外挂）

传入前几项，输出低n项的值

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <string>
 7 #include <map>
 8 #include <set>
 9 #include <cassert>
10 using namespace std;
11 #define rep(i,a,n) for (int i=a;i<n;i++)
12 #define per(i,a,n) for (int i=n-1;i>=a;i--)
13 #define pb push_back
14 #define mp make_pair
15 #define all(x) (x).begin(),(x).end()
16 #define fi first
17 #define se second
18 #define SZ(x) ((int)(x).size())
19 typedef vector<int> VI;
20 typedef long long ll;
21 typedef pair<int,int> PII;
22 const ll mod=1000000007;
23 ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
24 // head
25
26 int _,n;
27 namespace linear_seq {
28     const int N=10010;
29     ll res[N],base[N],_c[N],_md[N];
30
31     vector<int> Md;
32     void mul(ll *a,ll *b,int k) {
33         rep(i,0,k+k) _c[i]=0;
34         rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
35         for (int i=k+k-1;i>=k;i--) if (_c[i])
36             rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
37         rep(i,0,k) a[i]=_c[i];
38     }
39     int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
40 //        printf("%d\n",SZ(b));
41         ll ans=0,pnt=0;
42         int k=SZ(a);
43         assert(SZ(a)==SZ(b));
44         rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
45         Md.clear();
46         rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
47         rep(i,0,k) res[i]=base[i]=0;
48         res[0]=1;
49         while ((1ll<<pnt)<=n) pnt++;
50         for (int p=pnt;p>=0;p--) {
51             mul(res,res,k);
52             if ((n>>p)&1) {
53                 for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
54                 rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
55             }
56         }
57         rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
58         if (ans<0) ans+=mod;
59         return ans;
60     }
61     VI BM(VI s) {
62         VI C(1,1),B(1,1);
63         int L=0,m=1,b=1;
64         rep(n,0,SZ(s)) {
65             ll d=0;
66             rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
67             if (d==0) ++m;
68             else if (2*L<=n) {
69                 VI T=C;
70                 ll c=mod-d*powmod(b,mod-2)%mod;
71                 while (SZ(C)<SZ(B)+m) C.pb(0);
72                 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
73                 L=n+1-L; B=T; b=d; m=1;
74             } else {
75                 ll c=mod-d*powmod(b,mod-2)%mod;
76                 while (SZ(C)<SZ(B)+m) C.pb(0);
77                 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
78                 ++m;
79             }
80         }
81         return C;
82     }
83     int gao(VI a,ll n) {
84         VI c=BM(a);
85         c.erase(c.begin());
86         rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
87         return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
88     }
89 };
90
91 int main(){
92     //for (scanf("%d",&_);_;_--) {
93         while(scanf("%d",&n)!=EOF){
94             printf("%d\n",linear_seq::gao(VI{1,2,4,7,13,24},n-1));
95         }
96     //}
97 }