计蒜客蒜头君的兔子（神奇的杜教板子）

雨点打透心脏的1/2处 2022-05-15 11:47 337阅读 0赞

[题目传送门][Link 1]  
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));
        }
    }

[Link 1]: https://nanti.jisuanke.com/t/16442