ABC152-E Flatten

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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");
            WillReturn.Add("2 3 4");
            //13
        }
        else if (InputPattern == "Input2") {
            WillReturn.Add("5");
            WillReturn.Add("12 12 12 12 12");
            //5
        }
        else if (InputPattern == "Input3") {
            WillReturn.Add("3");
            WillReturn.Add("1000000 999999 999998");
            //996989508
        }
        else {
            string wkStr;
            while ((wkStr = Console.ReadLine()) != null) WillReturn.Add(wkStr);
        }
        return WillReturn;
    }

    const long Hou = 1000000007;

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

        var EntireSoinsuuDict = new Dictionary<long, long>();
        foreach (long EachA in AArr) {
            Dictionary<long, long> CurrSoinsuuDict = DeriveSoinsuuDict(EachA);

            foreach (var EachPair in CurrSoinsuuDict) {
                if (EntireSoinsuuDict.ContainsKey(EachPair.Key) == false) {
                    EntireSoinsuuDict[EachPair.Key] = EachPair.Value;
                }
                else {
                    long MaxCnt = EntireSoinsuuDict[EachPair.Key];
                    MaxCnt = Math.Max(MaxCnt, EachPair.Value);
                    EntireSoinsuuDict[EachPair.Key] = MaxCnt;
                }
            }
        }

        long LCM = 1;
        foreach (var EachPair in EntireSoinsuuDict) {
            long CurrVal = DeriveBekijyou(EachPair.Key, EachPair.Value, Hou);
            LCM *= CurrVal;
            LCM %= Hou;
        }

        long Answer = 0;
        foreach (long EachA in AArr) {
            Answer += LCM * DeriveGyakugen(EachA);
            Answer %= Hou;
        }
        Console.WriteLine(Answer);
    }

    // 素因数分解し、指数[素数]なDictを返す
    static Dictionary<long, long> DeriveSoinsuuDict(long pTarget)
    {
        var WillReturn = new Dictionary<long, long>();

        long CurrVal = pTarget;
        // ルートより大きい数を、素因素に持つとしても、1つだけのため
        // ルートまで試し割りを行い、素因数が残っていたら、追加する
        for (long I = 2; I * I <= pTarget; I++) {
            if (CurrVal % I > 0) continue;

            WillReturn[I] = 0;
            while (CurrVal % I == 0) {
                WillReturn[I]++;
                CurrVal /= I;
            }
            if (CurrVal == 1) break;
        }

        // 素因数が残っている場合
        if (CurrVal > 1) {
            WillReturn[CurrVal] = 1;
        }
        return WillReturn;
    }

    // 引数の逆元を求める
    static long DeriveGyakugen(long pLong)
    {
        return 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;
        }
    }
}

解説

考察すると
配列AのLCMを求め、
LCMを配列Aの各値で割った値の総和が、解と分かります。