题意：
小a终于放假了，它想在假期中去一些地方游玩，现在有N个景点，编号为
，同时小b也想出去游玩。由于一些特殊♂原因，他们的旅行计划必须满足一些条件
首先，他们可以从这N个景点中任意选几个游玩
设小a选出的景点集合为A，小b选的景点集合为B，则需要满足
1. A,B的交集不能为空集
2. A,B不能相互包含(A=B也属于相互包含)
注意：在这里我们认为(A,B)是无序的，即(A,B)和(B,A)是同一种方案
思路：
这道题如果手推的思路是这样的。先枚举A的个数的种类，然后枚举从A中选定几个为共有的个数，最后枚举B的个数的种类。
（注意图中标红处为-1，原来答案错了）
然后经过拆解，就可以得到答案。
//#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000") //c++ // #pragma GCC diagnostic error "-std=c++11" // #pragma comment(linker, "/stack:200000000") // #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") // #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3) #include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n' #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) #define max3(a,b,c) max(max(a,b), c); //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e8+7; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } /*-----------------------showtime----------------------*/ ll ksm(ll a,ll b){ ll res = 1; while(b > 0){ if(b & 1) res = res * a % mod; b >>= 1; a = a * a % mod; } return res % mod; } int main(){ ll n; cin>>n; ll ans = ksm(2, 2ll*n-1) % mod + 3ll * ksm(2, n-1)%mod - ((ksm(3,n+1) + 1) %mod *ksm(2,mod-2) % mod) ; cout<<(ans+mod)%mod<<endl; return 0; }