牛客-小a的旅行计划 + 数学推导

小a的旅行计划

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。

 

$$ \frac{-3 * 3^n + 4^n - 1}{2} + 3 * 2^{n-1}$$ 

 

// #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;
}