ACM-ICPC 2018 焦作赛区网络预赛 L Poor God Water（矩阵快速幂，BM）

https://nanti.jisuanke.com/t/31721

题意

有肉，鱼，巧克力三种食物，有几种禁忌，对于连续的三个食物：1.这三个食物不能都相同；2.若三种食物都有的情况，巧克力不能在中间；3.如果两边是巧克力，中间不能是肉或鱼。求方案数。

分析

将meat ,  chocolate,fish 用 0 ,1 , 2 表示。

对于 n 来说，我们只关注后两位，因为 若 n - 1 的所有方案解决的话，我们在 n - 1 的方案添加0, 1,2 就行，然后根据禁忌 去除不可能的方案。

 我们根据次状态 来更新现状态，然后矩阵快速幂。最后得到的矩阵的总和就是答案了。

另外，暴力推前十几项，然后用BM居然也能过！！黑科技黑科技

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5+10;
const int mod = 1e9 + 7;
const int N = 9;
struct Matrix{
    ll a[N][N];
    Matrix(){
        for(int i=0;i<N;i++)
            for(int j=0;j<N;j++)
                a[i][j]=0;
    }
};
Matrix mul(Matrix x,Matrix y){
    Matrix res;
    for(int i=0;i<N;i++)
        for(int j=0;j<N;j++)
            for(int k=0;k<N;k++)
                res.a[i][j]=(res.a[i][j]+x.a[i][k]*y.a[k][j]%mod)%mod;
    return res;
}
Matrix qpow(Matrix a,ll b){
    Matrix res;
    for(int i=0;i<N;i++) res.a[i][i]=1;
    while(b){
        if(b&1) res=mul(res,a);
        b>>=1;
        a=mul(a,a);
    }
    return res;
}
int A[9][9]={
    0,0,0,1,0,0,1,0,0,
    1,0,0,0,0,0,1,0,0,
    1,0,0,1,0,0,1,0,0,
    0,1,0,0,1,0,0,0,0,
    0,1,0,0,0,0,0,1,0,
    0,0,0,0,1,0,0,1,0,
    0,0,1,0,0,1,0,0,1,
    0,0,1,0,0,0,0,0,1,
    0,0,1,0,0,1,0,0,0
};
int main(){
    Matrix tmp;
    for(int i=0;i<N;i++)
        for(int j=0;j<N;j++)
            tmp.a[i][j]=A[i][j];
    int t;
    ll n;
    scanf("%d",&t);
    while(t--){
        scanf("%lld",&n);
        if(n==1) puts("3");
        else if(n==2) puts("9");
        else{
            Matrix ans;
            ans=qpow(tmp,n-2);
            ll res=0;
            for(int i=0;i<N;i++)
                for(int j=0;j<N;j++)
                    res=(res+ans.a[i][j])%mod;
            printf("%lld\n",res);
        }
    }
    return 0;
}

附上杜教BM模板。解决任何线性递推式

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
#include<bits/stdc++.h>
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 _,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() {
    int T;
    ll n;
    cin>>T;

    vector<int>v;
    v.push_back(3);
    v.push_back(9);
    v.push_back(20);
    v.push_back(46);
    v.push_back(106);
    v.push_back(244);
    v.push_back(560);
    v.push_back(1286);
    v.push_back(2956);
    v.push_back(6794);
    v.push_back(15610);
    v.push_back(35866);
    v.push_back(82416);
    while (T--) {
        scanf("%lld",&n);
        printf("%d\n",linear_seq::gao(v,n-1));
    }
}