Problem: Example

Input

1 10 2
1 1 1


Output

2.0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define ll long long
#define INF (1LL << 60)
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto& (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define sar(a,n) cout<<#a<<":";rep(pachico,n)cout<<" "<<a[pachico];cout<<endl
#define svec(v) cout<<#v<<":";rep(pachico,v.size())cout<<" "<<v[pachico];cout<<endl
#define svecp(v) cout<<#v<<":";each(pachico,v)cout<<" {"<<pachico.first<<":"<<pachico.second<<"}";cout<<endl
#define sset(s) cout<<#s<<":";each(pachico,s)cout<<" "<<pachico;cout<<endl
#define smap(m) cout<<#m<<":";each(pachico,m)cout<<" {"<<pachico.first<<":"<<pachico.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 100005;

ll f[1001];
ll dp1[1001],dp2[1001];
int mn[1001];

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int a,b,K;
    cin >> a >> b >> K;
    vl t(K+1);
    rep(i,K+1){
        cin >> t[i];
    }
    rep(i,b-a+1){
        f[i] = INF;
    }
    f[1] = 0;
    srep(i,2,b-a+1){
        rep(j,i+1){
            mn[j] = 0, dp1[j] = INF;
        }
        dp1[0] = 0;
        rep(j,K+1){
            rep(k,i+1){
                dp2[k] = INF;
            }
            srep(k,j+1,i+1){
                srep(l,max(mn[k],j),k){
                    if(dp2[k] > dp1[l]+(k-l)*t[j]+f[k-l]){
                        dp2[k] = dp1[l]+(k-l)*t[j]+f[k-l];
                        mn[k] = l;
                    }
                }
            }
            if(j) cmn(f[i],dp2[i]);
            swap(dp1,dp2);
        }
    }
    printf("%.12lf\n",(double)f[b-a]/(b-a));
    return 0;
}