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题目传送门
下面的代码别问我为什么这么写,我也不会,只是听说能解决一切线性递推式
只要把前面几项先处理出来就可以了,一般到50项应该就没问题了
代码:
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long LL;
const LL mod=1000000007;
LL mypow(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 maxn=10010;
LL res[maxn],base[maxn],c[maxn],md[maxn];
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<<pnt)<=n) pnt++;
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*mypow(b,mod-2)%mod;
while (SZ(C)<SZ(B)+m) C.pb(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*mypow(b,mod-2)%mod;
while (SZ(C)<SZ(B)+m) C.pb(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(){
LL num[15];
while(scanf("%d",&n)==1){
vector<int>v;
memset(num,0,sizeof(num));
num[1]=1;
for(int i=1;i<=20;i++){
LL cnt=0;
for(int j=0;j<10;j++) cnt+=num[j],cnt%=mod;
v.pb(cnt);
for(int j=10;j>=1;j--) num[j]=num[j-1];num[0]=0;
for(int j=2;j<=10;j++) num[0]+=num[j],num[0]%=mod;
num[10]=0;
}
printf("%d\n",linear_seq::gao(v,n-1));
}
}