https://blog.csdn.net/qq_36876305/article/details/80275708
1 #include <bits/stdc++.h>
2
3 using namespace std;
4 #define rep(i,a,n) for (long long i=a;i<n;i++)
5 #define per(i,a,n) for (long long i=n-1;i>=a;i--)
6 #define pb push_back
7 #define mp make_pair
8 #define all(x) (x).begin(),(x).end()
9 #define fi first
10 #define se second
11 #define SZ(x) ((long long)(x).size())
12 typedef vector<
long long>
VI;
13 typedef
long long ll;
14 typedef pair<
long long,
long long>
PII;
15 const ll mod=1e9+
7;
16 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;}
17 // head
18
19 long long _,n;
20 namespace linear_seq
21 {
22 const long long N=
10010;
23 ll res[N],
base[N],_c[N],_md[N];
24
25 vector<
long long>
Md;
26 void mul(ll *a,ll *b,
long long k)
27 {
28 rep(i,
0,k+k) _c[i]=
0;
29 rep(i,
0,k)
if (a[i]) rep(j,
0,k)
30 _c[i+j]=(_c[i+j]+a[i]*b[j])%
mod;
31 for (
long long i=k+k-
1;i>=k;i--)
if (_c[i])
32 rep(j,
0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%
mod;
33 rep(i,
0,k) a[i]=
_c[i];
34 }
35 long long solve(ll n,VI a,VI b)
36 {
// a 系数 b 初值 b[n+1]=a[0]*b[n]+...
37 // printf("%d\n",SZ(b));
38 ll ans=
0,pnt=
0;
39 long long k=
SZ(a);
40 assert(SZ(a)==
SZ(b));
41 rep(i,
0,k) _md[k-
1-i]=-a[i];_md[k]=
1;
42 Md.clear();
43 rep(i,
0,k)
if (_md[i]!=
0) Md.push_back(i);
44 rep(i,
0,k) res[i]=
base[i]=
0;
45 res[
0]=
1;
46 while ((1ll<<pnt)<=n) pnt++
;
47 for (
long long p=pnt;p>=
0;p--
)
48 {
49 mul(res,res,k);
50 if ((n>>p)&
1)
51 {
52 for (
long long i=k-
1;i>=
0;i--) res[i+
1]=res[i];res[
0]=
0;
53 rep(j,
0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%
mod;
54 }
55 }
56 rep(i,
0,k) ans=(ans+res[i]*b[i])%
mod;
57 if (ans<
0) ans+=
mod;
58 return ans;
59 }
60 VI BM(VI s)
61 {
62 VI C(
1,
1),B(
1,
1);
63 long long L=
0,m=
1,b=
1;
64 rep(n,
0,SZ(s))
65 {
66 ll d=
0;
67 rep(i,
0,L+
1) d=(d+(ll)C[i]*s[n-i])%
mod;
68 if (d==
0) ++
m;
69 else if (
2*L<=
n)
70 {
71 VI T=
C;
72 ll c=mod-d*powmod(b,mod-
2)%
mod;
73 while (SZ(C)<SZ(B)+m) C.pb(
0);
74 rep(i,
0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%
mod;
75 L=n+
1-L; B=T; b=d; m=
1;
76 }
77 else
78 {
79 ll c=mod-d*powmod(b,mod-
2)%
mod;
80 while (SZ(C)<SZ(B)+m) C.pb(
0);
81 rep(i,
0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%
mod;
82 ++
m;
83 }
84 }
85 return C;
86 }
87 long long gao(VI a,ll n)
88 {
89 VI c=
BM(a);
90 c.erase(c.begin());
91 rep(i,
0,SZ(c)) c[i]=(mod-c[i])%
mod;
92 return solve(n,c,VI(a.begin(),a.begin()+
SZ(c)));
93 }
94 };
95
96 int main()
97 {
98 while(~scanf(
"%I64d", &
n))
99 { printf(
"%I64d\n",linear_seq::gao(VI{
1,
5,
11,
36,
95,
281,
781,
2245,
6336,
18061,
51205},n-
1));
100 }
101 }
转载于:https://www.cnblogs.com/MekakuCityActor/p/9655047.html
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