小a的旅行計劃

題意：

小a終於放假了，它想在假期中去一些地方遊玩，現在有N個景點，編號為 ，同時小b也想出去遊玩。由於一些特殊♂原因，他們的旅行計劃必須滿足一些條件 首先，他們可以從這N個景點中任意選幾個遊玩  設小a選出的景點集合為A，小b選的景點集合為B，則需要滿足  1. A,B的交集不能為空集  2. A,B不能相互包含(A=B也屬於相互包含)  注意：在這裡我們認為(A,B)是無序的，即(A,B)和(B,A)是同一種方案   思路： 　　 這道題如果手推的思路是這樣的。先列舉A的個數的種類，然後列舉從A中選定幾個為共有的個數，最後列舉B的個數的種類。   （注意圖中標紅處為-1，原來答案錯了） 然後經過拆解，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;//這是一個大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//這是一個小根堆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)  //用來壓行
#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;    //黃金分割點
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;
}

View Code