牛客-小a的旅行计划 + 数学推导

时间：2024-08-13 22:05:26

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，编号为牛客-小a的旅行计划 + 数学推导，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。
 牛客-小a的旅行计划 + 数学推导
（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。
牛客-小a的旅行计划 + 数学推导
//#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;  //

const ll nmos = 0x80000000;  //-2147483648

const int inf = 0x3f3f3f3f;

const ll inff = 0x3f3f3f3f3f3f3f3f; //

const int mod = 1e8+;

const double esp = 1e-;

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=;int f=;char ch=getchar();

    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

    return x=f?-x:x;

}

/*-----------------------showtime----------------------*/

            ll ksm(ll a,ll b){

                ll res = ;

                while(b > ){

                    if(b & ) res = res * a % mod;

                    b >>= ;

                    a = a * a % mod;

                }

                return res % mod;

            }

int main(){

            ll n;   cin>>n;

            ll ans = ksm(, 2ll*n-) % mod + 3ll * ksm(, n-)%mod - ((ksm(,n+) + ) %mod *ksm(,mod-) % mod) ;

            cout<<(ans+mod)%mod<<endl;

            return ;

}

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