ABC257-E Addition and Multiplication 2

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("5");
            WillReturn.Add("5 4 3 3 2 5 3 5 3");
            //95
        }
        else if (InputPattern == "Input2") {
            WillReturn.Add("20");
            WillReturn.Add("1 1 1 1 1 1 1 1 1");
            //99999999999999999999
        }
        else {
            string wkStr;
            while ((wkStr = Console.ReadLine()) != null) WillReturn.Add(wkStr);
        }
        return WillReturn;
    }

    static void Main()
    {
        List<string> InputList = GetInputList();
        int N = int.Parse(InputList[0]);
        int[] CArr = InputList[1].Split(' ').Select(pX => int.Parse(pX)).ToArray();

        int MinCost = CArr.Min();
        char BaseChar = '?';
        for (int I = CArr.GetUpperBound(0); 0 <= I; I--) {
            if (CArr[I] == MinCost) {
                BaseChar = (I + 1).ToString()[0];
                break;
            }
        }

        string AnswerStr = new string(BaseChar, N / MinCost);
        int Rest = N % MinCost;

        char[] AnswerArr = AnswerStr.ToCharArray();

        for (int I = 0; I <= AnswerArr.GetUpperBound(0); I++) {
            for (int J = CArr.GetUpperBound(0); 0 <= J; J--) {
                char CurrChar = (J + 1).ToString()[0];
                if (CurrChar <= BaseChar) break;

                int NeedCost = CArr[J] - MinCost;
                if (NeedCost <= Rest) {
                    AnswerArr[I] = CurrChar;
                    Rest -= NeedCost;
                    break;
                }
            }
            if (Rest <= 0) break;
        }

        Console.WriteLine(new string(AnswerArr.ToArray()));
    }
}

解説

桁数を最優先で増やすのが最適なので
まずは、コストが最小の中で最大の数字を
可能な限り並べます。

次に、余った金額で
上位桁から順に、
9,8,7,6,5,4,3,2の優先順位で変更可能かを見てます。