ACM-ICPC 2018 焦作赛区网络预赛 L. Poor God Water（杜教）

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题目链接：https://nanti.jisuanke.com/t/31721

样例输入 
3
3
4
15
样例输出 
20
46
435170

题意：三种食物，肉、鱼、巧克力，每小时吃一种，要求不能存在三种情况：连续三小时吃同一种食物；连续三小时，吃三种不同食物，巧克力在中间吃；连续三小时，第一小时和第三小时吃巧克力。问有几种符合条件的吃法

思路：卡了很久这题，最后抱着试一试的心态，掏出了珍藏已久的杜教模板，过了。机房大佬以前写过一个板子，可以求出递推式，跑了一下，是f(n) = 2*f(n-1) - 1*f(n-2) + 4*f(n-3) + f(n-4) 

这个是杜教：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#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;}
// head

int _;
ll 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) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
//        printf("%d\n",SZ(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*powmod(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*powmod(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() {
    for (scanf("%d",&_);_;_--) {
        scanf("%lld",&n);
        printf("%d\n",linear_seq::gao(VI{3,9,20,46,106,244,560,1286,2956,6794},n-1));//放前几项
    }
    return 0;
}

这个是机房大佬的可以出递推式的代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,a,n) for(int i=a;i<n;i++)
namespace linear
{
    ll mo=1000000009;
    vector<ll> v;
    double a[105][105],del;
    int k;
    struct matrix
    {
        int n;
        ll a[50][50];
        matrix operator * (const matrix & b)const
        {
            matrix c;
            c.n=n;
            rep(i,0,n)rep(j,0,n)c.a[i][j]=0;
            rep(i,0,n)rep(j,0,n)rep(k,0,n)
            c.a[i][j]=(c.a[i][j]+a[i][k]*b.a[k][j]%mo)%mo;
            return c;
        }
    }A;
    bool solve(int n)
    {
        rep(i,1,n+1)
        {
            int t=i;
            rep(j,i+1,n+1)if(fabs(a[j][i])>fabs(a[t][i]))t=j;
            if(fabs(del=a[t][i])<1e-6)return false;
            rep(j,i,n+2)swap(a[i][j],a[t][j]);
            rep(j,i,n+2)a[i][j]/=del;
            rep(t,1,n+1)if(t!=i)
            {
                del=a[t][i];
                rep(j,i,n+2)a[t][j]-=a[i][j]*del;
            }
        }
        return true;
    }
    void build(vector<ll> V)
    {
        v=V;
        int n=(v.size()-1)/2;
        k=n;
        while(1)
        {
            rep(i,0,k+1)
            {
                rep(j,0,k)a[i+1][j+1]=v[n-1+i-j];
                a[i+1][k+1]=1;
                a[i+1][k+2]=v[n+i];
            }
            if(solve(n+1))break;
            n--;k--;
        }
        A.n=k+1;
        rep(i,0,A.n)rep(j,0,A.n)A.a[i][j]=0;
        rep(i,0,A.n)A.a[i][0]=(int)round(a[i+1][A.n+1]);
        rep(i,0,A.n-2)A.a[i][i+1]=1;
        A.a[A.n-1][A.n-1]=1;
    }
    void formula()
    {
        printf("f(n) =");
        rep(i,0,A.n-1)printf(" (%lld)*f(n-%d) +",A.a[i][0],i+1);
        printf(" (%lld)\n",A.a[A.n-1][0]);
    }
    ll cal(ll n)
    {
        if(n<v.size())return v[n];
        n=n-k+1;
        matrix B,T=A;
        B.n=A.n;
        rep(i,0,B.n)rep(j,0,B.n)B.a[i][j]=i==j?1:0;
        while(n)
        {
            if(n&1)B=B*T;
            n>>=1;
            T=T*T;
        }
        ll ans=0;
        rep(i,0,B.n-1)ans=(ans+v[B.n-2-i]*B.a[i][0]%mo)%mo;
        ans=(ans+B.a[B.n-1][0])%mo;
        while(ans<0)ans+=mo;
        return ans;
    }
}

int main()
{
//  vector<ll> V={1 ,4 ,9 ,16,25,36,49};
//  vector<ll> V={1 ,1 ,2 ,3 ,5 ,8 ,13};
//  vector<ll> V={2 ,2 ,3 ,4 ,6 ,9 ,14};
    vector<ll> V={3,9,20,46,106,244,560,1286,2956,6794};
    linear::build(V);
    linear::formula();
    ll n;
    while(~scanf("%lld",&n))
    {
        printf("%lld\n",linear::cal(n-1));
    }
    return 0;
}