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

 int n;
 LL dp[maxn];

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

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

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

 int n;
 int dp[maxn][];

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

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=;
 ll powmod(ll a,ll b) {
     ll res=;
     a%=mod;
     assert(b>=);
     for(; b; b>>=) {
         if(b&)res=res*a%mod;
         a=a*a%mod;
     }
     return res;
 }
 // head
 int _,n;
 namespace linear_seq {
 const int N=;
 ll res[N],base[N],_c[N],_md[N];
 vector<int> Md;
 void mul(ll *a,ll *b,int k) {
     rep(i,,k+k) _c[i]=;
     rep(i,,k) if (a[i]) rep(j,,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
     for (int i=k+k-; i>=k; i--) if (_c[i])
             rep(j,,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
     rep(i,,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=,pnt=;
     int k=SZ(a);
     assert(SZ(a)==SZ(b));
     rep(i,,k) _md[k--i]=-a[i];
     _md[k]=;
     Md.clear();
     rep(i,,k) if (_md[i]!=) Md.push_back(i);
     rep(i,,k) res[i]=base[i]=;
     res[]=;
     while ((1ll<<pnt)<=n) pnt++;
     for (int p=pnt; p>=; p--) {
         mul(res,res,k);
         if ((n>>p)&) {
             for (int i=k-; i>=; i--) res[i+]=res[i];
             res[]=;
             rep(j,,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
         }
     }
     rep(i,,k) ans=(ans+res[i]*b[i])%mod;
     if (ans<) ans+=mod;
     return ans;
 }
 VI BM(VI s) {
     VI C(,),B(,);
     int L=,m=,b=;
     rep(n,,SZ(s)) {
         ll d=;
         rep(i,,L+) d=(d+(ll)C[i]*s[n-i])%mod;
         if (d==) ++m;
         else if (*L<=n) {
             VI T=C;
             ll c=mod-d*powmod(b,mod-)%mod;
             while (SZ(C)<SZ(B)+m) C.pb();
             rep(i,,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
             L=n+-L,B=T,b=d,m=;
         } else {
             ll c=mod-d*powmod(b,mod-)%mod;
             while (SZ(C)<SZ(B)+m) C.pb();
             rep(i,,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,,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{,,,,,,,,,},n-));
 }

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）：