ABC350-E Toward 0

問題へのリンク

C#のソース

using System;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static string InputPattern = "InputX";

    static List<string> GetInputList()
    {
        var WillReturn = new List<string>();

        if (InputPattern == "Input1") {
            WillReturn.Add("3 2 10 20");
            //20.000000000000000
        }
        else if (InputPattern == "Input2") {
            WillReturn.Add("3 2 20 20");
            //32.000000000000000
        }
        else if (InputPattern == "Input3") {
            WillReturn.Add("314159265358979323 4 223606797 173205080");
            //6418410657.7408381
        }
        else {
            string wkStr;
            while ((wkStr = Console.ReadLine()) != null) WillReturn.Add(wkStr);
        }
        return WillReturn;
    }

    static long mA;
    static long mX;
    static long mY;

    static void Main()
    {
        List<string> InputList = GetInputList();
        long[] wkArr = InputList[0].Split(' ').Select(pX => long.Parse(pX)).ToArray();

        long CurrVal = wkArr[0];
        mA = wkArr[1];
        mX = wkArr[2];
        mY = wkArr[3];

        decimal Result = DFS(CurrVal);
        Console.WriteLine(Result);
    }

    // 現在値を引数として、残り金額の期待値を返す
    static Dictionary<long, decimal> mMemo = new Dictionary<long, decimal>();
    static decimal DFS(long CurrVal)
    {
        if (mMemo.ContainsKey(CurrVal)) {
            return mMemo[CurrVal];
        }

        if (CurrVal == 0) return 0;

        var KouhoList = new List<decimal>();

        // X円払う場合
        KouhoList.Add(mX + DFS(CurrVal / mA));

        // Y円払う場合

        // 当たりを引くまでの期待値
        decimal Ex1 = 0;
        Ex1 += mY * 6M / 5M;

        decimal Ex2 = 0;
        for (long I = 2; I <= 6; I++) {
            Ex2 += 1M / 5M * DFS(CurrVal / I);
        }
        decimal NewEx = Ex1 + Ex2;
        KouhoList.Add(NewEx);

        return mMemo[CurrVal] = KouhoList.Min();
    }
}

解説

メモ化再帰で解いてます。