线性递推式模板 hdu6198为例
今天打2017沈阳网络赛的时候，第五题好像是个找规律的题（因为我好像找出规律了，但是没写），队友直接一个模版，把我推出的前几项放进去后直接就可以把后面的弄出来。。。太强了
写这篇博客保存一下这个强无敌的模板，可以解决任何线性递推式.。这个板子是我一个学长从百度之星复赛上扒的杜教的板子。所以我们现在要做的是用推出递推式的前几项.，然后扔进这个板子就可以了。
mod换成题目的就OK，前几项丢的越多越好， 一般8个是绝对可以推出来的，个别的少一点也行。
模板
#include
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
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;}
int _,n;
namespace linear_seq {
const int N=10010;
ll res[N],base[N],_c[N],_md[N];
vector<int> Md;
void mul(ll *a,ll *b,int k) {
rep(i,0,k+k) _c[i]=0;
rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
for (int i=k+k-1;i>=k;i--) if (_c[i])
rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
rep(i,0,k) a[i]=_c[i];
}
int solve(ll n,VI a,VI b) {
ll ans=0,pnt=0;
int k=SZ(a);
assert(SZ(a)==SZ(b));
rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
Md.clear();
rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
rep(i,0,k) res[i]=base[i]=0;
res[0]=1;
while ((1ll<for (int p=pnt;p>=0;p--) {
mul(res,res,k);
if ((n>>p)&1) {
for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
}
}
rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
if (ans<0) ans+=mod;
return ans;
}
VI BM(VI s) {
VI C(1,1),B(1,1);
int L=0,m=1,b=1;
rep(n,0,SZ(s)) {
ll d=0;
rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
if (d==0) ++m;
else if (2*L<=n) {
VI T=C;
ll c=mod-d*powmod(b,mod-2)%mod;
while (SZ(C)0);
rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
L=n+1-L; B=T; b=d; m=1;
} else {
ll c=mod-d*powmod(b,mod-2)%mod;
while (SZ(C)0);
rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
++m;
}
}
return C;
}
int gao(VI a,ll n) {
VI c=BM(a);
c.erase(c.begin());
rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
}
};
int main() {
for (scanf("%d",&_);_;_--) {
scanf("%d",&n);
printf("%d\n",linear_seq::gao(VI{2,24,96,416,1536,5504,18944,64000,212992,702464},n-1));
}
}
hdu6198
#include
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=998244353;
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;
}
int _,n;
namespace linear_seq {
const int N=10010;
ll res[N],base[N],_c[N],_md[N];
vector<int> Md;
void mul(ll *a,ll *b,int k) {
for(int i = 0 ; i < k + k ; ++i)
_c[i]=0;
for(int i = 0 ; i < k ;++i)
if (a[i])
for(int j = 0 ;j < k ;++ j)
_c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
for (int i=k+k-1;i>=k;i--)
if (_c[i])
for(int j = 0 ; j<(int ) Md.size() ; ++ j)
_c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
for(int i =0 ; i< k ; ++i)
a[i]=_c[i];
}
int solve(ll n,VI a,VI b) {
ll ans=0,pnt=0;
int k=SZ(a);
assert( SZ(a) == SZ(b) );
for(int i = 0 ;i < k ; ++ i)
_md[k-1-i] = -a[i] ; _md[k] = 1 ;
Md.clear() ;
for(int i =0 ; i < k ; ++ i)
if (_md[i]!=0)
Md.push_back(i);
for(int i = 0; i< k ;++ i)
res[i]=base[i]=0;
res[0]=1;
while ((1ll<for (int p=pnt;p>=0;p--) {
mul(res,res,k);
if ((n>>p)&1) {
for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
for(int j = 0 ;j < (int)Md.size() ; ++ j)
res[ Md[j] ]=(res[ Md[j] ]-res[k]*_md[Md[j]])%mod;
}
}
rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
if (ans<0) ans+=mod;
return ans;
}
VI BM(VI s) {
VI C(1,1),B(1,1);
int L=0,m=1,b=1;
for(int n= 0 ;n < (int)s.size(); ++ n ) {
ll d=0;
// rep(i,0,L+1)
for(int i =0 ; i < L +1 ;++ i)
d=(d+(ll)C[i]*s[n-i])%mod;
if (d==0) ++m;
else if (2*L<=n) {
VI T=C;
ll c=mod-d*powmod(b,mod-2)%mod;
while (SZ(C)0);
// rep(i,0,SZ(B))
for(int i =0 ; i < (int)B.size(); ++ i)
C[i+m]=(C[i+m]+c*B[i])%mod;
L=n+1-L; B=T; b=d; m=1;
} else {
ll c=mod-d*powmod(b,mod-2)%mod;
while (SZ(C)0);
for(int i = 0 ;i <(int) B.size() ; ++ i)
C[i+m]=(C[i+m]+c*B[i])%mod;
++m;
}
}
return C;
}
ll gao(VI a,ll n) {
VI c=BM(a);
c.erase(c.begin());
for( int i = 0 ; i < (int)c.size( );++i )
c[i]=(mod-c[i])%mod;
return (ll)solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
}
};
int main() {
long long ___ ;VI a;
int N,v;
a.push_back(4);
a.push_back(12);
a.push_back(33);
a.push_back(88);
a.push_back(232);
a.push_back(609);
a.push_back(1596);
for (;~scanf("%lld",&___);)
printf("%lld\n",linear_seq::gao(a,___-1));
return 0 ;
}