HDU - 1099 - Lottery - 概率dp

http://acm.hdu.edu.cn/showproblem.php?pid=1099

最最简单的概率dp，完全是等概率转移。
设dp[i]为已有i张票，还需要抽几次才能集齐的期望。

那么dp[n]=0，因为我们已经集齐了。

\[dp[i]=(\frac{i}{n}*dp[i]+\frac{n-i}{n}*dp[i+1])+1 \]

移项得答案。

然后写个分数类，注意约分。

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

struct Fraction{
    ll fz;
    ll fm;

    Fraction(){}

    Fraction(ll z,ll m):fz(z),fm(m){}

    Fraction operator+(Fraction f){
        ll m=fm/__gcd(fm,f.fm)*f.fm;
        Fraction res;
        res.fz=fz*(m/fm)+f.fz*(m/f.fm);
        res.fm=m;
        res.tf();
        return res;
    }

    Fraction operator-(Fraction f){
        ll m=fm/__gcd(fm,f.fm)*f.fm;
        Fraction res;
        res.fz=fz*(m/fm)-f.fz*(m/f.fm);
        res.fm=m;
        res.tf();
        return res;
    }

    Fraction operator*(Fraction f){
        Fraction res;
        res.fz=fz*f.fz;
        res.fm=fm*f.fm;
        res.tf();
        return res;
    }

    void tf(){
        ll g=__gcd(fz,fm);
        fz/=g;
        fm/=g;
    }



    void show(){
        ll zs=fz/fm;
        ll ys=fz%fm;

        if(ys==0){
            cout<<zs<<endl;
        }
        else{
            string l1,l2,l3;
            l1=" ";
            l2=" ";
            l3=" ";
            while(zs){
                l2=char(zs%10+'0')+l2;
                zs/=10;
                l1+=" ";
                l3+=" ";
            }

            ll cfm=fm;

            string n3="";
            string n2="";
            while(cfm){
                n3=char(cfm%10+'0')+n3;
                cfm/=10;
                n2+="-";
            }

            l3+=n3;
            l2+=n2;

            string n1="";

            ll cfz=ys;

            while(cfz){
                n1=char(cfz%10+'0')+n1;
                cfz/=10;
            }

            l1+=n1;

            cout<<l1<<endl<<l2<<endl<<l3<<endl;
        }
    }
};

Fraction dp[23];

int main(){
    int n;
    while(cin>>n){
        dp[n]=Fraction(0,1);
        for(int i=n-1;i>=0;i--){
            dp[i]=dp[i+1]+Fraction(n,n-i);
        }
        dp[0].show();
    }
}