Problem:
Score : 800 points

Problem Statement

Given are integers N and X. For each integer k between 0 and X (inclusive), find the answer to the following question, then compute the sum of all those answers, modulo 998244353.

Let us repeat the following operation on the integer k. Operation: if the integer is currently odd, subtract 1 from it and divide it by 2; otherwise, divide it by 2 and add 2^{N-1} to it. How many operations need to be performed until k returns to its original value? (The answer is considered to be 0 if k never returns to its original value.)

Constraints

1 \leq N \leq 2\times 10^5

0 \leq X < 2^N

X is given in binary and has exactly N digits. (In case X has less than N digits, it is given with leading zeroes.)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X

Output

Print the sum of the answers to the questions for the integers between 0 and X (inclusive), modulo 998244353.

Sample Input 1

3
111

Sample Output 1

40

For example, for k=3, the operation changes k as follows: 1,0,4,6,7,3. Therefore the answer for k=3 is 6.

Sample Input 2

6
110101

Sample Output 2

616

Sample Input 3

30
001110011011011101010111011100

Sample Output 3

549320998

Solution in C# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using CompLib.Mathematics;
using CompLib.Util;

public class Program
{
    private int N;
    private bool[] X;

    public void Solve()
    {
        var sc = new Scanner();
        N = sc.NextInt();
        X = sc.Next().Select(c => c == '1').ToArray();

        /*
         * 奇数 1桁ずらす
         * 偶数 1桁ずらす、一番上1追加
         *
         * 一番下の桁を反転して一番上に
         * 2週すれば必ず元に戻る
         *
         * 例1
         * 0,1,3,4,6,7 6回
         * 2,5         2回
         * 6*6 + 2*2 = 40
         *
         * 
         * 
         * 2m回で戻る 長さmの部分列、反転が交互に
         * mはNの約数
         * 
         */

        List<int> div = new List<int>();
        {
            List<int> r = new List<int>();
            for (int i = 1; i * i <= N; i++)
            {
                if (N % i == 0)
                {
                    div.Add(i);
                    if (i != N / i) r.Add(N / i);
                }
            }

            for (int i = r.Count - 1; i >= 0; i--)
            {
                div.Add(r[i]);
            }
        }

        ModInt[] cnt = new ModInt[div.Count];
        for (int i = 0; i < div.Count; i++)
        {
            // i桁2進 反転... 交互
            // 最大
            // 上位div[i]桁そのままができるか?
            int d = div[i];
            if ((N / d) % 2 == 0) continue;
            bool f = true;
            for (int j = d; j < N; j++)
            {
                int dd = j / d;
                int mm = j % d;

                if ((X[mm] ^ (dd % 2 == 1)) && !X[j])
                {
                    f = false;
                    break;
                }

                if ((X[mm] ^ (dd % 2 == 0)) && X[j])
                {
                    break;
                }
            }

            ModInt p = 0;
            for (int j = 0; j < d; j++)
            {
                p *= 2;
                if (X[j]) p += 1;
            }

            if (!f)
            {
                // できない
                // 上位d桁-1
                p -= 1;
            }

            // 交互にp個できる
            cnt[i] += p + 1;
            for (int j = i + 1; j < div.Count; j++)
            {
                if (div[j] % d == 0 && (div[j] / d) % 2 == 1)
                {
                    cnt[j] -= cnt[i];
                }
            }
        }

        ModInt ans = 0;
        for (int i = 0; i < div.Count; i++)
        {
            ans += 2 * div[i] * cnt[i];
        }

        Console.WriteLine(ans);
    }

    public static void Main(string[] args) => new Program().Solve();
}

// https://bitbucket.org/camypaper/complib
namespace CompLib.Mathematics
{
    #region ModInt

    /// <summary>
    /// [0,<see cref="Mod"/>) までの値を取るような数
    /// </summary>
    public struct ModInt
    {
        /// <summary>
        /// 剰余を取る値．
        /// </summary>
        // public const long Mod = (int) 1e9 + 7;
        public const long Mod = 998244353;

        /// <summary>
        /// 実際の数値．
        /// </summary>
        public long num;

        /// <summary>
        /// 値が <paramref name="n"/> であるようなインスタンスを構築します．
        /// </summary>
        /// <param name="n">インスタンスが持つ値</param>
        /// <remarks>パフォーマンスの問題上，コンストラクタ内では剰余を取りません．そのため，<paramref name="n"/> ∈ [0,<see cref="Mod"/>) を満たすような <paramref name="n"/> を渡してください．このコンストラクタは O(1) で実行されます．</remarks>
        public ModInt(long n)
        {
            num = n;
        }

        /// <summary>
        /// このインスタンスの数値を文字列に変換します．
        /// </summary>
        /// <returns>[0,<see cref="Mod"/>) の範囲内の整数を 10 進表記したもの．</returns>
        public override string ToString()
        {
            return num.ToString();
        }

        public static ModInt operator +(ModInt l, ModInt r)
        {
            l.num += r.num;
            if (l.num >= Mod) l.num -= Mod;
            return l;
        }

        public static ModInt operator -(ModInt l, ModInt r)
        {
            l.num -= r.num;
            if (l.num < 0) l.num += Mod;
            return l;
        }

        public static ModInt operator *(ModInt l, ModInt r)
        {
            return new ModInt(l.num * r.num % Mod);
        }

        public static implicit operator ModInt(long n)
        {
            n %= Mod;
            if (n < 0) n += Mod;
            return new ModInt(n);
        }

        /// <summary>
        /// 与えられた 2 つの数値からべき剰余を計算します．
        /// </summary>
        /// <param name="v">べき乗の底</param>
        /// <param name="k">べき指数</param>
        /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
        public static ModInt Pow(ModInt v, long k)
        {
            return Pow(v.num, k);
        }

        /// <summary>
        /// 与えられた 2 つの数値からべき剰余を計算します．
        /// </summary>
        /// <param name="v">べき乗の底</param>
        /// <param name="k">べき指数</param>
        /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
        public static ModInt Pow(long v, long k)
        {
            long ret = 1;
            for (k %= Mod - 1; k > 0; k >>= 1, v = v * v % Mod)
                if ((k & 1) == 1)
                    ret = ret * v % Mod;
            return new ModInt(ret);
        }

        /// <summary>
        /// 与えられた数の逆元を計算します．
        /// </summary>
        /// <param name="v">逆元を取る対象となる数</param>
        /// <returns>逆元となるような値</returns>
        /// <remarks>法が素数であることを仮定して，フェルマーの小定理に従って逆元を O(log N) で計算します．</remarks>
        public static ModInt Inverse(ModInt v)
        {
            return Pow(v, Mod - 2);
        }
    }

    #endregion

    #region Binomial Coefficient

    public class BinomialCoefficient
    {
        public ModInt[] fact, ifact;

        public BinomialCoefficient(int n)
        {
            fact = new ModInt[n + 1];
            ifact = new ModInt[n + 1];
            fact[0] = 1;
            for (int i = 1; i <= n; i++)
                fact[i] = fact[i - 1] * i;
            ifact[n] = ModInt.Inverse(fact[n]);
            for (int i = n - 1; i >= 0; i--)
                ifact[i] = ifact[i + 1] * (i + 1);
            ifact[0] = ifact[1];
        }

        public ModInt this[int n, int r]
        {
            get
            {
                if (n < 0 || n >= fact.Length || r < 0 || r > n) return 0;
                return fact[n] * ifact[n - r] * ifact[r];
            }
        }

        public ModInt RepeatedCombination(int n, int k)
        {
            if (k == 0) return 1;
            return this[n + k - 1, k];
        }
    }

    #endregion
}


namespace CompLib.Util
{
    using System;
    using System.Linq;

    class Scanner
    {
        private string[] _line;
        private int _index;
        private const char Separator = ' ';

        public Scanner()
        {
            _line = new string[0];
            _index = 0;
        }

        public string Next()
        {
            if (_index >= _line.Length)
            {
                string s;
                do
                {
                    s = Console.ReadLine();
                } while (s.Length == 0);

                _line = s.Split(Separator);
                _index = 0;
            }

            return _line[_index++];
        }

        public string ReadLine()
        {
            _index = _line.Length;
            return Console.ReadLine();
        }

        public int NextInt() => int.Parse(Next());
        public long NextLong() => long.Parse(Next());
        public double NextDouble() => double.Parse(Next());
        public decimal NextDecimal() => decimal.Parse(Next());
        public char NextChar() => Next()[0];
        public char[] NextCharArray() => Next().ToCharArray();

        public string[] Array()
        {
            string s = Console.ReadLine();
            _line = s.Length == 0 ? new string[0] : s.Split(Separator);
            _index = _line.Length;
            return _line;
        }

        public int[] IntArray() => Array().Select(int.Parse).ToArray();
        public long[] LongArray() => Array().Select(long.Parse).ToArray();
        public double[] DoubleArray() => Array().Select(double.Parse).ToArray();
        public decimal[] DecimalArray() => Array().Select(decimal.Parse).ToArray();
    }
}