#lightoj #cp #problem_solving #dp #expected_value #probability
Idea
Expected Value
- The expectation at
i is E(i) = (E(i+1) + E(i+2) + E(i+3) + E(i+4) + E(i+5) + E(i+6)) / cnt + gold[i]
where cnt is how many of E(i+x) is valid
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#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define PII pair<int,int>
#define PLL pair<LL,LL>
#define MP make_pair
#define F first
#define S second
#define INF INT_MAX
inline void optimizeIO()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
const int nmax = 1e2+7;
const LL LINF = 1e17;
string to_str(LL x)
{
stringstream ss;
ss<<x;
return ss.str();
}
long double ara[103];
long double dp[103];
int n;
long double solve(int pos)
{
if(pos==n-1)
return ara[pos];
long double &ret = dp[pos];
if(ret>=0.0)
return ret;
ret = 0;
int cnt = 0;
for(int i=1; i<=6; i++)
{
if(pos+i<n)
{
ret += ara[pos] + solve(pos+i);
}
else
cnt++;
}
ret /= (6.0-cnt);
return ret;
}
int main()
{
optimizeIO();
int tc;
cin>>tc;
for(int q=1; q<=tc; q++)
{
cin>>n;
for(int i=0; i<n; i++)
cin>>ara[i];
memset(dp,-1,sizeof dp);
long double ans = solve(0);
cout << fixed << showpoint;
cout << setprecision(10);
cout<<"Case "<<q<<": "<<ans<<endl;
}
return 0;
}
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