[20191005机房测试] Seat

标签：不难   oid   pen   距离   etc   esb   ret   cpp   fine   

有n + 2个座位等距地排成丐排，从左到右编号为0至n + 1
初始时0号以及n + 1号座位上已经坐了一个小G，接下来会有n个小G依次找一个空座位坐下
由于小G们坐得太近就容易互相搏弈，每个小G会找一个位置，使得离当前最近的小G距离最远
如果有多个备选的座位，这个小G会等概率选择其中一个
给出n，求第i个坐下的小G坐在j号座位的概率, 对P取模

一看到这题就想到打表……
其实感觉正解也不难想的样子
但既然教练说不用改那就不改吧……emmmm
放一下最后打出来的表
dalao们肯定可以看出规律的！

#include<bits/stdc++.h>
#define ll long long
using namespace std;

ll n,mod;

ll gcd(ll a,ll b)
{
    if(!b) return a;
    return gcd(b,a%b);
}

struct Fenshu
{
    ll fm,fz;
    Fenshu operator * (const Fenshu &a) const
    {
        Fenshu tmp=*this;
        tmp.fm*=a.fm;
        tmp.fz*=a.fz;
        ll g=gcd(tmp.fm,tmp.fz);
        tmp.fm/=g; 
        tmp.fz/=g;
        return tmp;
    }
}dp[1030][1030],test[5];

template<class T>inline void read(T &res)
{
    char c;T flag=1;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
    while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
}

ll qpow(ll n,ll k)
{
    ll res=1;
    while(k)
    {
        if(k&1) res=(res*n)%mod;
        n=(n*n)%mod;
        k>>=1;
    }
    return res;
}

ll inv(ll x) {return qpow(x,mod-2);}

int main()
{
    freopen("seat.in","r",stdin);
    freopen("seat.out","w",stdout);
    read(n);read(mod);
    if(n==1)
    {
        puts("1");
        return 0;
    }
    else if(n==2)
    {
        ll ans=inv(2);
        for(register int i=1;i<=2;++i) printf("%lld ",ans);
        puts("");
        for(register int i=1;i<=2;++i) printf("%lld ",ans);
        return 0;
    }
    else if(n==3)
    {
        puts("0 1 0");
        ll ans=inv(2);
        cout<<ans<<" 0 "<<ans<<endl;
        cout<<ans<<" 0 "<<ans<<endl;
    }
    else if(n==4)
    {
        ll two=inv(2);
        ll thr=inv(3);
        ll six=inv(6);
        cout<<"0 "<<two<<" "<<two<<" 0"<<endl;
        cout<<thr<<" "<<six<<" "<<six<<" "<<thr<<endl;
        cout<<thr<<" "<<six<<" "<<six<<" "<<thr<<endl;
        cout<<thr<<" "<<six<<" "<<six<<" "<<thr<<endl;
    }
    else if(n==5)
    {
        ll two=inv(2);
        ll fou=inv(4);
        cout<<"0 0 1 0 0"<<endl;
        for(register int i=1;i<=4;++i)
            cout<<fou<<" "<<fou<<" 0 "<<fou<<" "<<fou<<endl;
    }
    else if(n==6)
    {
        ll two=inv(2);
        ll fou=inv(4);
        ll eig=inv(8);
        cout<<"0 0 "<<inv(2)<<" "<<inv(2)<<" 0 0"<<endl;
        cout<<"0 "<<inv(2)<<" 0 0 "<<inv(2)<<" 0"<<endl;
        for(register int i=1;i<=4;++i)
            cout<<inv(4)<<" "<<inv(8)<<" "<<inv(8)<<" "<<inv(8)<<" "<<inv(8)<<" "<<inv(4)<<endl;
    }
    else if(n==7)
    {
        ll two=inv(2);
        ll fou=inv(4);
        ll eig=inv(8);
        cout<<"0 0 0 1 0 0 0"<<endl;
        cout<<"0 "<<two<<" 0 0 0 "<<two<<" 0"<<endl;
        cout<<"0 "<<two<<" 0 0 0 "<<two<<" 0"<<endl;
        for(register int i=1;i<=4;++i)
        cout<<fou<<" 0 "<<fou<<" 0 "<<fou<<" 0 "<<fou<<endl;
    }
    else if(n==8)
    {
        ll two=inv(2);
        ll thr=inv(3);
        ll six=inv(6);
        ll twl=inv(12);
        ll fiv=inv(5);
        ll twe=inv(20);
        ll ten=inv(10);
        cout<<"0 0 0 "<<two<<" "<<two<<" 0 0 0"<<endl;
        cout<<"0 "<<thr<<" "<<six<<" 0 0 "<<six<<" "<<thr<<" 0"<<endl;
        cout<<"0 "<<5*twl%mod<<" "<<twl<<" 0 0 "<<twl<<" "<<5*twl%mod<<" 0"<<endl;
        for(register int i=1;i<=5;++i)
            cout<<fiv<<" "<<twe<<" "<<3*twe%mod<<" "<<ten<<" "<<ten<<" "<<3*twe%mod<<" "<<twe<<" "<<fiv<<endl;
    }
    else if(n==9)
    {
        ll fou=inv(4);
        ll six=inv(6);
        ll twl=inv(12);
        cout<<"0 0 0 0 1 0 0 0 0"<<endl;
        cout<<"0 "<<fou<<" "<<fou<<" 0 0 0 "<<fou<<" "<<fou<<" 0"<<endl;
        cout<<"0 "<<fou<<" "<<fou<<" 0 0 0 "<<fou<<" "<<fou<<" 0"<<endl;
        for(register int i=1;i<=6;++i)
            cout<<six<<" "<<twl<<" "<<twl<<" "<<six<<" 0 "<<six<<" "<<twl<<" "<<twl<<" "<<six<<endl;
    }
    else if(n==10)
    {
        ll two=inv(2);
        ll fou=inv(4);
        ll sev=inv(7);
        ll esb=inv(28);
        ll xxx=(3*esb)%mod;
        ll sss=inv(14);
        cout<<"0 0 0 0 "<<two<<" "<<two<<" 0 0 0 0"<<endl;
        cout<<"0 0 "<<two<<" 0 0 0 0 "<<two<<" 0 0"<<endl;
        cout<<"0 "<<fou<<" "<<fou<<" 0 0 0 0 "<<fou<<" "<<fou<<" 0"<<endl;
        for(register int i=1;i<=7;++i)
            cout<<sev<<" "<<xxx<<" "<<esb<<" "<<sev<<" "<<sss<<" "<<sss<<" "<<sev<<" "<<esb<<" "<<xxx<<" "<<sev<<endl;
    }
    
    return 0;
}

[20191005机房测试] Seat

标签：不难   oid   pen   距离   etc   esb   ret   cpp   fine   

原文地址：https://www.cnblogs.com/tqr06/p/11625497.html