Dog Card is a card game. In the game, there are a total of 2n cards in the deck, each card has a value, and the values of these 2n cards form a permutation of 1 ~ 2n. There is a skill that works as follows: 1. Draw a card from the top of the deck. 2. If the deck is empty, then skip to step 3, otherwise you guess whether the card on the top of the deck has a higher value than your last drawn card and draw a card from the top of the deck. If your guess is correct, then repeat this step, otherwise skip to step 3. 3. End this process. Nana enjoys playing this game, although she may not be skilled at it. Therefore, her guessing strategy when using this skill is simple: if the value of the last drawn card is less than or equal to n, then she guesses that the next oard's valve is higher, ther wse, she guedses thet the next card's vaue s lomler she wârns tb dmokt tor anfafrhlm decks of cards (Obviously, there are (2n)! cases), how many cards she can draw in total if she uses the skill only once in each case. Since this number can be very large,please provide the answer modulo a given value.Dog Card is a card game. In the game, there are a total of 2n cards in the deck, each card has a value, and the values of these 2n cards form a permutation of 1 ~ 2n. There is a skill that works as follows: 1. Draw a card from the top of the deck. 2. If the deck is empty, then skip to step 3, otherwise you guess whether the card on the top of the deck has a higher value than your last drawn card and draw a card from the top of the deck. If your guess is correct, then repeat this step, otherwise skip to step 3. 3. End this process. Nana enjoys playing this game, although she may not be skilled at it. Therefore, her guessing strategy when using this skill is simple: if the value of the last drawn card is less than or equal to n, then she guesses that the next oard's valve is higher, ther wse, she guedses thet the next card's vaue s lomler she wârns tb dmokt tor anfafrhlm decks of cards (Obviously, there are (2n)! cases), how many cards she can draw in total if she uses the skill only once in each case. Since this number can be very large,please provide the answer modulo a given value.给出c++代码及中文解释

中文解释：

这个问题可以使用动态规划来解决。我们定义一个二维数组dp，其中dp[i][j]表示当剩余卡牌数量为i时，上一张抽到的卡牌值为j时，能够抽取的最大卡牌数量。

根据题目要求，我们可以得到以下状态转移方程：
- 如果j > n，即上一张抽到的卡牌值大于n，那么下一张卡牌的值可以是1到2n之间的任意值，并且抽取的卡牌数量加1。
- 如果j <= n，即上一张抽到的卡牌值小于等于n，那么下一张卡牌的值只能是j+1到2n之间的值，并且抽取的卡牌数量加1。

最终，我们需要计算所有剩余卡牌数量为2n时，所有可能的上一张抽到的卡牌值j的累加和。

C++代码如下所示：

#include <iostream>
#include <vector>

using namespace std;

int dogCard(int n, int mod) {
    vector<vector<int>> dp(2 * n + 1, vector<int>(2 * n + 1, 0));
    
    for (int i = 1; i <= 2 * n; i++) {
        dp[1][i] = 1;
    }
    
    for (int i = 2; i <= 2 * n; i++) {
        for (int j = 1; j <= 2 * n; j++) {
            if (j > n) {
                dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % mod;
            } else {
                dp[i][j] = (dp[i][j] + dp[i - 1][j + 1]) % mod;
            }
        }
    }
    
    int result = 0;
    for (int j = 1; j <= 2 * n; j++) {
        result = (result + dp[2 * n][j]) % mod;
    }
    
    return result;
}

int main() {
    int n, mod;
    cout << "Enter the value of n: ";
    cin >> n;
    cout << "Enter the value of mod: ";
    cin >> mod;
    
    int result = dogCard(n, mod);
    cout << "Total cards that can be drawn: " << result << endl;
    
    return 0;
}

在主函数中，我们从用户输入获取n和mod的值，然后调用dogCard函数来计算可以抽取的总卡牌数量，并输出结果。

注意：由于结果可能非常大，我们需要对结果取模mod。