【ZOJ 3640 Help Me Escape】概率dp

ZOJ3640
题意 有个人啊 他有初始的攻击力$f$ 总共有$n$个怪物 每个怪物呢都有$c[i]$,$t[i]$
其中$t[i]$ = $(1+sqrt(5))/2\ast c[i]\ast c[i]$ 他每天随机挑选一个怪物来Pk 如果这个人打得过怪物呢 他会花$t[i]$天逃出去 如果打不过那么他会把自己的攻击力变成$f+c[i]$ 问你逃出去的期望
直接求这个次数期望是不好求的
转换个思路 我们设$dp[f]$代表$f$攻击下的期望天数
那么对我们来说 如果$f>c[i]$那么我们直接加上$t[i]$
如果小于 那么我们记忆化搜索$dp[f]+=dp[f+c[i]+1$
因为是每天随机挑选一个 所以最后要处以$n$ 代表等概率的pk

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
#define P ((1+sqrt(5))/2.0)
const int MAX_N = 20025;
double dp[MAX_N];
int c[MAX_N],t[MAX_N],n;
double dfs(int f)
{
    if(dp[f]>0) return dp[f];
    double ans = 0;
    for(int i = 1;i<=n;++i)
    {
        if(f>c[i]) ans+=1.0*t[i];
        else ans+=dfs(f+c[i])+1;
    }
    dp[f] = ans/(1.0*n);return dp[f];
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int f;
    while(scanf("%d%d",&n,&f)==2)
    {
    memset(dp,0,sizeof(dp));
    for(int i = 1;i<=n;++i) scanf("%d",&c[i]),t[i] = c[i]*c[i]*P;
    dfs(f);
    printf("%.3f\n",dp[f]);
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}