ABC405-E Fruit Lineup

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("1 1 1 1");
            //5
        }
        else if (InputPattern == "Input2") {
            WillReturn.Add("1 2 4 8");
            //2211
        }
        else if (InputPattern == "Input3") {
            WillReturn.Add("834150 21994 467364 994225");
            //947921688
        }
        else {
            string wkStr;
            while ((wkStr = Console.ReadLine()) != null) WillReturn.Add(wkStr);
        }
        return WillReturn;
    }

    const long Hou = 998244353;

    static void Main()
    {
        List<string> InputList = GetInputList();
        long[] wkArr = InputList[0].Split(' ').Select(pX => long.Parse(pX)).ToArray();
        long A = wkArr[0];
        long B = wkArr[1];
        long C = wkArr[2];
        long D = wkArr[3];

        long Answer = 0;

        ChooseMod InsChooseMod = new ChooseMod(A + B + C + D, Hou);

        // 1番右のAの位置を全探索
        for (long I = A; I <= A + B; I++) {
            long TotalLen = A + B;
            long RightLen = TotalLen - I;
            long LeftLen = I - 1;

            long RestA = A - 1;
            long RestB = B - RightLen;

            long Pattern1 = InsChooseMod.DeriveChoose(LeftLen, RestB);

            // 一番右のAより左にCを配置
            long MoveCnt = RightLen + D;
            long Pattern2 = InsChooseMod.DeriveChoose(MoveCnt + C, C);

            long CurrAnswer = Pattern1 * Pattern2;
            //Console.WriteLine("1番右が{0}だと{1}通り", I, CurrAnswer);

            CurrAnswer %= Hou;
            Answer += CurrAnswer;
            Answer %= Hou;
        }
        Console.WriteLine(Answer);
    }

    // nCr (mod Hou)を求める
    static long DeriveChoose(long pN, long pR)
    {
        if (pN < pR) return 0;

        pR = Math.Min(pR, pN - pR);

        long WillReturn = 1;
        for (long I = pN - pR + 1; I <= pN; I++) {
            WillReturn *= I;
            WillReturn %= Hou;
        }
        for (long I = 2; I <= pR; I++) {
            WillReturn *= DeriveGyakugen(I);
            WillReturn %= Hou;
        }
        return WillReturn;
    }

    // 引数の逆元を求める
    static Dictionary<long, long> mMemoGyakugen = new Dictionary<long, long>();
    static long DeriveGyakugen(long pLong)
    {
        if (mMemoGyakugen.ContainsKey(pLong)) {
            return mMemoGyakugen[pLong];
        }
        return mMemoGyakugen[pLong] = DeriveBekijyou(pLong, Hou - 2, Hou);
    }

    // 繰り返し2乗法で、(NのP乗) Mod Mを求める
    static long DeriveBekijyou(long pN, long pP, long pM)
    {
        long CurrJyousuu = pN % pM;
        long CurrShisuu = 1;
        long WillReturn = 1;

        while (true) {
            // 対象ビットが立っている場合
            if ((pP & CurrShisuu) > 0) {
                WillReturn = (WillReturn * CurrJyousuu) % pM;
            }

            CurrShisuu *= 2;
            if (CurrShisuu > pP) return WillReturn;
            CurrJyousuu = (CurrJyousuu * CurrJyousuu) % pM;
        }
    }
}

#region ChooseMod
// 二項係数クラス
internal class ChooseMod
{
    private long mHou;

    private long[] mFacArr;
    private long[] mFacInvArr;
    private long[] mInvArr;

    // コンストラクタ
    internal ChooseMod(long pCnt, long pHou)
    {
        mHou = pHou;
        mFacArr = new long[pCnt + 1];
        mFacInvArr = new long[pCnt + 1];
        mInvArr = new long[pCnt + 1];

        mFacArr[0] = mFacArr[1] = 1;
        mFacInvArr[0] = mFacInvArr[1] = 1;
        mInvArr[1] = 1;
        for (int I = 2; I <= pCnt; I++) {
            mFacArr[I] = mFacArr[I - 1] * I % mHou;
            mInvArr[I] = mHou - mInvArr[mHou % I] * (mHou / I) % mHou;
            mFacInvArr[I] = mFacInvArr[I - 1] * mInvArr[I] % mHou;
        }
    }

    // nCrを返す
    internal long DeriveChoose(long pN, long pR)
    {
        if (pN < pR) return 0;
        if (pN < 0 || pR < 0) return 0;
        return mFacArr[pN] * (mFacInvArr[pR] * mFacInvArr[pN - pR] % mHou) % mHou;
    }
}
#endregion

解説

以下の手順で解いてます。

手順1
一番右のAの位置を全探索しつつ、
AとBだけを配置

手順2
Dを配置

手順3
Cを、一番右のA以降に配置