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Time Limit: 20 Sec
Memory Limit: 256 MB
http://codeforces.com/gym/100187/problem/D
Everyone knows that the battle of Endor is just a myth fabled by George Lucas for promotion of his movie. Actually, no battle of Endor has happened and the First Galactic Empire prospers to this day.
There are creatures of n races living in the First Galactic Empire. In order to demonstrate their freedom, equality and brotherhood the Emperor commanded to introduce the holidays. During each of these holidays creatures of one non-empty subset of races should give gifts to creatures of another non-empty subset of races, not intersecting the first one.
The Emperor‘s stuff is not very strong in maths so you should calculate how many such holidays can be introduced. Two holidays are considered different if they differ in the subset of races which give gifts or in the subset of races which receive gifts.
Input
The input contains the only integer n (1 ≤ n ≤ 200000) — the number of races living in the First Galactic Empire.
Output
Find the number of holidays the Emperor commanded to introduce. This number can be very large, so output the reminder of division of this number by 109 + 9.
Sample Input
2
Sample Output
2
题意
每一个集合都是一个种族，这个种族会给任何与他没有相交的集合礼物，有n个人，问你最后要给几个礼物？
题解：
数学题啦，推公式推公式，排列组合就好了
代码
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define test freopen("test.txt","r",stdin) #define maxn 2001001 #define mod 1000000009 #define eps 1e-9 const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } //************************************************************************************** ll ans[maxn]; void pre() { ans[0]=1; for(int i=1;i<maxn;i++) ans[i]=ans[i-1]*2,ans[i]%=mod; } ll fac[maxn]; ll qpow(ll a,ll b) { ll ans=1;a%=mod; for(ll i=b;i;i>>=1,a=a*a%mod) if(i&1)ans=ans*a%mod; return ans; } ll C(ll n,ll m) { if(m>n||m<0)return 0; ll s1=fac[n],s2=fac[n-m]*fac[m]%mod; return s1*qpow(s2,mod-2)%mod; } int main() { fac[0]=1; for(int i=1;i<maxn;i++) fac[i]=fac[i-1]*i%mod; pre(); int n=read(); ll ans1=0; for(int i=1;i<=n;i++) { ans1+=(ans[n-i]-1)*C(n,i); ans1%=mod; } cout<<ans1<<endl; }
Codeforces Gym 100187D D. Holidays 排列组合
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原文地址：http://www.cnblogs.com/qscqesze/p/4657529.html
