ABC467-C Adjacent Sums (easy)

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");
            WillReturn.Add("1 1 1");
            WillReturn.Add("1 1");
            //1
        }
        else if (InputPattern == "Input2") {
            WillReturn.Add("2 2");
            WillReturn.Add("1 1");
            WillReturn.Add("0");
            //0
        }
        else if (InputPattern == "Input3") {
            WillReturn.Add("10 2");
            WillReturn.Add("0 0 0 1 1 0 1 0 1 0");
            WillReturn.Add("0 1 0 1 0 1 0 1 0");
            //4
        }
        else {
            string wkStr;
            while ((wkStr = Console.ReadLine()) != null) WillReturn.Add(wkStr);
        }
        return WillReturn;
    }

    static long[] GetSplitArr(string pStr)
    {
        return (pStr == "" ? new string[0] : pStr.Split(' ')).Select(pX => long.Parse(pX)).ToArray();
    }

    static long[] mAArr;
    static long[] mBArr;

    static void Main()
    {
        List<string> InputList = GetInputList();
        mAArr = GetSplitArr(InputList[1]);
        mBArr = GetSplitArr(InputList[2]);

        var CostList = new List<long>();

        long[] CurrAArr1 = (long[])mAArr.Clone();
        long Cost1 = DeriveCost(CurrAArr1);

        long[] CurrAArr2 = (long[])mAArr.Clone();
        CurrAArr2[0]++;
        long Cost2 = 1 + DeriveCost(CurrAArr2);

        Console.WriteLine(Cost1 < Cost2 ? Cost1 : Cost2);
    }

    // コストを返す
    static long DeriveCost(long[] pAArr)
    {
        long Cost = 0;
        for (long I = 1; I <= pAArr.GetUpperBound(0); I++) {
            long Sum = pAArr[I - 1] + pAArr[I];
            Sum %= 2;
            if (Sum != mBArr[I - 1]) {
                pAArr[I]++;
                Cost++;
            }
        }
        return Cost;
    }
}

解説

数列Aの初項を決めると
残りの項が決定するので

初項を0と1の2通りで、
総コストの低いほうが解になります。