Tetrahedron（Codeforces Round #113 (Div. 2) + 打表找规律 + dp计数）

题目链接：

　　https://codeforces.com/contest/166/problem/E

题目：

题意：

　　给你一个三菱锥，初始时你在D点，然后你每次可以往相邻的顶点移动，问你第n步回到D点的方案数。

思路：

　　打表找规律得到的序列是0,3,6,21,60,183,546,1641,4920,14763，通过肉眼看或者oeis可以得到规律为。

　　dp计数：dp[i][j]表示在第i步时站在位置j的方案数，j的取值为[0,3]，分别表示D,A,B,C点，转移方程肯定是从其他三个点转移。

代码实现如下：

 非dp计数：

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 1e7 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int n;
LL dp[maxn];

int qpow(int x, int n) {
    int res = 1;
    while(n) {
        if(n & 1) res = 1LL * res * x % mod;
        x = 1LL * x * x % mod;
        n >>= 1;
    }
    return res;
}

int main(){
    int inv = qpow(4, mod - 2);
    LL cnt = 1;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        cnt = cnt * 3 % mod;
        if(i & 1) dp[i] = (cnt - 3 + mod) % mod * inv % mod;
        else dp[i] = (cnt + 3) % mod * inv % mod;
    }
    printf("%lld\n", dp[n]);
    return 0;
}

dp计数代码：

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 1e7 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int n;
int dp[maxn][4];

int main(){
    scanf("%d", &n);
    dp[0][0] = 1;
    for(int i = 1; i <= n; i++) {
        for(int j = 0; j < 4; j++) {
            for(int k = 0; k < 4; k++) {
                if(j == k) continue;
                dp[i][j] = (1LL * dp[i][j] + dp[i-1][k]) % mod;
            }
        }
    }
    printf("%d\n", dp[n][0]);
    return 0;
}

BM代码：

#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 _,n;
namespace linear_seq {
const int N=100100;
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() {
    scanf("%d",&n);
    printf("%d\n",linear_seq::gao(VI{0,3,6,21,60,183,546,1641,4920,14763},n-1));
}

三份代码跑的时间如下（忽略MLE那发，从上到下分别为BM,DP，公式）：