Educational Codeforces Round 57 (Rated for Div. 2)G. Lucky Tickets【NTT】_all bus tickets in berland have their numbers. a n-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_42671946/article/details/91044138

本文探讨了在Berland的公交车票系统中，寻找所有可能的幸运车票数量的算法实现。幸运车票定义为票号前半部分数字之和等于后半部分数字之和的票。文章通过使用快速傅立叶变换(FFT)和快速幂运算，解决了当票号由特定数字组成且可能包含前导零的情况下的计算问题。

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G. Lucky Tickets

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

All bus tickets in Berland have their numbers. A number consists of nn digits (nn is even). Only kk decimal digits d1,d2,…,dkd1,d2,…,dk can be used to form ticket numbers. If 00 is among these digits, then numbers may have leading zeroes. For example, if n=4n=4 and only digits 00 and 44can be used, then 00000000, 40044004, 44404440 are valid ticket numbers, and 00020002, 0000, 4444344443 are not.

A ticket is lucky if the sum of first n/2n/2 digits is equal to the sum of remaining n/2n/2 digits.

Calculate the number of different lucky tickets in Berland. Since the answer may be big, print it modulo 998244353998244353.

Input

The first line contains two integers nn and kk (2≤n≤2⋅105,1≤k≤10)(2≤n≤2⋅105,1≤k≤10) — the number of digits in each ticket number, and the number of different decimal digits that may be used. nn is even.

The second line contains a sequence of pairwise distinct integers d1,d2,…,dkd1,d2,…,dk (0≤di≤9)(0≤di≤9) — the digits that may be used in ticket numbers. The digits are given in arbitrary order.

Output

Print the number of lucky ticket numbers, taken modulo 998244353998244353.

Examples

input

Copy

4 2
1 8

output

Copy

input

Copy

20 1
6

output

Copy

input

Copy

10 5
6 1 4 0 3

output

Copy

input

Copy

1000 7
5 4 0 1 8 3 2

output

Copy

460571165

Note

In the first example there are 66 lucky ticket numbers: 11111111, 18181818, 18811881, 81188118, 81818181 and 88888888.

There is only one ticket number in the second example, it consists of 2020 digits 66. This ticket number is lucky, so the answer is 11.

分析：容易想到用 $dp[i][j]$ 表示前i个凑成j的方案数，最后答案就是 $\sum _{i=0}^{maxi}dp[n/2][i]^2$

转移方程 $dp[i][j]=\sum _{k=0}^{9}dp[i-1][j-k]*d[k]$ ,d[k]表示k是否存在。

容易得到 $dp[0][i]=d[i]$ ，那么结果就是d[k]卷n/2次，直接NTT+快速幂优化即可。

#include "bits/stdc++.h"

using namespace std;
const int mod = 998244353;

long long qk(long long a, long long n) {
    long long ans = 1;
    while (n) {
        if (n & 1)ans = ans * a % mod;
        n >>= 1;
        a = a * a % mod;
    }
    return ans;
}

void ntt(long long A[], int n, int type) {
    for (int i = 0, j = 0; i < n; ++i) {
        if (i < j) swap(A[i], A[j]);
        for (int l = (n >> 1); (j ^= l) < l; l >>= 1);
    }
    for (int s = 2; s <= n; s <<= 1) {
        int t = (s >> 1);
        int u = (type == 1) ? (qk(3, (mod - 1) / s)) : (qk(3, (mod - 1) - (mod - 1) / s));
        for (int i = 0; i < n; i += s) {
            for (int p = 1, j = 0; j < t; ++j, p = 1LL * p * u % mod) {
                int x = A[i + j], y = 1LL * A[i + j + t] * p % mod;
                A[i + j] = (x + y) % mod, A[i + j + t] = (x + mod - y) % mod;
            }
        }
    }
    if (type == -1) {
        int invn = qk(n, mod - 2);
        for (int i = 0; i < n; ++i) {
            A[i] = 1LL * A[i] * invn % mod;
        }
    }
}

long long f[2000004];

int main() {
    int n, k, d, maxi = 0;
    memset(f, 0, sizeof(f));
    cin >> n >> k;
    for (int i = 0; i < k; ++i) {
        cin >> d;
        maxi = max(maxi, d);
        f[d] = 1;
    }
    int len = 1;
    while (len <= ((n / 2) * maxi))
    {
        len <<= 1;
    }

    ntt(f, len, 1);
    for (int i = 0; i < len; ++i) {
        f[i] = qk(f[i], n / 2);
    }
    ntt(f, len, -1);

    long long ans = 0;
    for (int i = 0; i < len; ++i) {
        ans = (ans + 1LL * f[i] * f[i] % mod) % mod;
    }
    cout << ans;
}