ABC293-E Geometric Progression

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 4 7");
            //5
        }
        else if (InputPattern == "Input2") {
            WillReturn.Add("8 10 9");
            //0
        }
        else if (InputPattern == "Input3") {
            WillReturn.Add("1000000000 1000000000000 998244353");
            //919667211
        }
        else {
            string wkStr;
            while ((wkStr = Console.ReadLine()) != null) WillReturn.Add(wkStr);
        }
        return WillReturn;
    }

    static void Main()
    {
        List<string> InputList = GetInputList();
        long[] wkArr = InputList[0].Split(' ').Select(pX => long.Parse(pX)).ToArray();
        long Kouhi = wkArr[0];
        long Kousuu = wkArr[1];
        long Hou = wkArr[2];

        long Result = DeriveTouhisuuretuSum(1, Kouhi, Kousuu, Hou);
        Console.WriteLine(Result);
    }

    // 初項,公比,項数,法を引数とし、等比数列の和を返す(法が逆元を持たなくても可)
    static long DeriveTouhisuuretuSum(long pSyokou, long pKouhi, long pKousuu, long pHou)
    {
        pSyokou %= pHou;

        // 項数が1の場合
        if (pKousuu == 1) {
            return pSyokou;
        }

        // 項数が奇数の場合
        if (pKousuu % 2 == 1) {
            long WillReturn = pSyokou;
            WillReturn += DeriveTouhisuuretuSum(pSyokou * pKouhi, pKouhi, pKousuu - 1, pHou);
            WillReturn %= pHou;
            return WillReturn;
        }

        // 項数が偶数の場合
        long SumLeft = DeriveTouhisuuretuSum(pSyokou, pKouhi, pKousuu / 2, pHou);
        SumLeft %= pHou;

        long SumRight = SumLeft * DeriveBekijyou(pKouhi, pKousuu / 2, pHou);
        SumRight %= pHou;

        return (SumLeft + SumRight) % pHou;
    }

    // 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

再帰で半分ずつに分割してます。