小a的旅行计划（组合数学与逆元）

• 注释：$(a+b{)}^n=\sum _{i=0}^n{a}^i{b}^{n-i}{C}_n^i$，本章用到$\{\begin{array}{l}a=1\\ b=1\end{array}$与$\{\begin{array}{l}a=2\\ b=1\end{array}$与$\{\begin{array}{l}a=1\\ b=2\end{array}$的特殊情况
• 题面
• 题意：见题面。
• 解决思路：由题意不难看出集合$A,B$的个数都不能为$1$或$n$$(1,n$的时候肯定相互包含$)$
  直接枚举加推导就好
  考虑$A$，先从$n$个景点挑出$i$个，即：$\sum _{i=2}^{n-1}{C}_n^i$
  再来考虑$B$，为了满足条件，只要$B$在$A$中挑出大于等于$1$个景点，并且在$A$的补集中挑出大于等于$1$个景点即可。
  注意：$B$不能把$A$中所有的元素挑出，不然会包含的。
  $B$在$A$中：$\sum _{j=1}^{i-1}{C}_i^j$
  $B$在$A$的补集中：$\sum _{k=1}^{n-i}{C}_{n-i}^k$
  推导
  $\therefore \sum _{i=2}^{n-1}{C}_n^i\sum _{j=1}^{i-1}{C}_i^j\sum _{k=1}^{n-i}{C}_{n-i}^k$
  $=\sum _{i=2}^{n-1}{C}_n^i\sum _{j=1}^{i-1}{C}_i^j(\sum _{k=0}^{n-i}{C}_{n-i}^k-C_{n-i}^0)$
  $=\sum _{i=2}^{n-1}{C}_n^i\sum _{j=1}^{i-1}{C}_i^j(2^{n-i}-1)$
  $=\sum _{i=2}^{n-1}{C}_n^i(\sum _{j=0}^i{C}_i^j-C_i^0-C_i^i)(2^{n-i}-1)$
  $=\sum _{i=2}^{n-1}{C}_n^i(2^i-2)(2^{n-i}-1)$
  $=\sum _{i=2}^{n-1}{C}_n^i(2^n-2^i-2\times 2^{n-i}+2)$
  $=\sum _{i=2}^{n-1}{C}_n^i(2^n+2-2^i-2\times 2^{n-i})$
  $=(2^n+2)\sum _{i=2}^{n-1}{C}_n^i-\sum _{i=2}^{n-1}{2}^i{C}_n^i-2\sum _{i=2}^{n-1}{2}^{n-i}{C}_n^i$
  $=(2^n+2)(\sum _{i=0}^n{C}_n^i-C_n^0-C_n^1-C_n^n)-(\sum _{i=0}^n{2}^i{C}_n^i-2^0{C}_n^0-2^1{C}_n^1-2^n{C}_n^n)-2(\sum _{i=0}^n{2}^{n-i}{C}_n^i-2^n{C}_n^0-2^{n-1}{C}_n^1-2^0{C}_n^n)$
  $=(2^n+2)(2^n-n-2)-(3^n-1-2n-2^n)-2(3^n-2^n-n{2}^{n-1}-1)$
  $=4^n-1-3\times 3^n+3\times 2^n$
  由于$(A,B)，(B,A)$是同一种方案
  $\therefore ans=\frac{4^n-1-3\times 3^n+3\times 2^n}{2}$
  $ans=\frac{4^n-1-3\times 3^n}{2}+3\times 2^{n-1}$
  分母直接快速幂求解逆元就好
• AC代码

//优化
#pragma GCC optimize(2)
//C
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
//C++
#include<unordered_map>
#include<algorithm>
#include<iostream>
#include<istream>
#include<iomanip>
#include<climits>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
//宏定义
#define N 1010
#define DoIdo main
//#define scanf scanf_s
#define it set<ll>::iterator
#define TT template<class T>
//定义+命名空间
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e8 + 7;
const ll INF = 1e18;
const int maxn = 1e6 + 10;
using namespace std;
//全局变量
//函数区
ll max(ll a, ll b) {
    
     return a > b ? a : b; }
ll min(ll a, ll b) {
    
     return a < b ? a : b; }
ll quick_pow(ll a, ll b) {
    
    
	ll res = 1; a %= mod;
	while (b) {
    
    
		if (b & 1) {
    
    
			res = (res * a) % mod;
		}
		a = (a * a) % mod;
		b >>= 1;
	}
	return res;
}
//主函数
int DoIdo() {
    
    

	ios::sync_with_stdio(false);
	cin.tie(NULL), cout.tie(NULL);

	ll n;
	cin >> n;

	ll ans2 = (3 * quick_pow(2, n - 1)) % mod;
	ll ans1 = ((quick_pow(4, n) - 1 - quick_pow(3, n + 1)) % mod + mod) * quick_pow(2, mod - 2);

	cout << (ans1 + ans2) % mod;
	return 0;
}
//分割线---------------------------------QWQ
/*



*/