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.给出中文解释和c++代码
时间: 2024-04-23 17:27:56 浏览: 23
Dog Card是一种纸牌游戏。游戏中,牌堆里共有2n张牌,每张牌都有一个值,这2n张牌的值形成了1~2n的排列。
游戏中有一个技能,其工作方式如下:
1. 从牌堆顶抽一张牌。
2. 如果牌堆为空,则跳到步骤3;否则,你猜测牌堆顶的牌是否比你上次抽到的牌的值大,并且抽一张牌。如果你的猜测正确,则重复这一步骤,否则跳到步骤3。
3. 结束这个过程。
Nana喜欢玩这个游戏,虽然她可能不太擅长。因此,她在使用这个技能时的猜测策略很简单:如果上次抽到的牌的值小于等于n,她猜测下一张牌的值更大;否则,她猜测下一张牌的值更小。她想知道,在所有可能的牌堆情况下(显然有(2n)!种情况),如果她只使用这个技能一次,她最多能抽多少张牌。由于这个数字可能非常大,请给出对给定值取模后的答案。
以下是一个C++代码示例,用于计算Nana能够抽取的最多牌数(取模后):
```cpp
#include <iostream>
using namespace std;
const int MOD = 1000000007;
const int MAXN = 2005;
int dp[MAXN][MAXN]; // dp[i][j]表示前i张牌中有j张牌被猜对了的情况下的最多抽牌数
int dogCard(int n) {
// 初始化dp数组
for (int i = 0; i <= 2 * n; i++) {
for (int j = 0; j <= n; j++) {
dp[i][j] = 0;
}
}
dp[0][0] = 1; // 初始化边界条件
for (int i = 1; i <= 2 * n; i++) {
for (int j = 0; j <= n; j++) {
// 上一张牌的值小于等于n,猜测下一张牌更大
if (i - j <= n) {
dp[i][j] += dp[i - 1][j] * (i - j);
}
// 上一张牌的值大于n,猜测下一张牌更小
if (j > 0) {
dp[i][j] += dp[i - 1][j - 1] * j;
}
// 取模运算
dp[i][j] %= MOD;
}
}
int ans = 0;
for (int j = 0; j <= n; j++) {
ans = (ans + dp[2 * n][j]) % MOD;
}
return ans;
}
int main() {
int n;
cin >> n;
int result = dogCard(n);
cout << result << endl;
return 0;
}
```
在这段代码中,使用动态规划的思想,通过填表的方式计算出dp数组。最后,将dp[2n][j](0 <= j <= n)的值求和,即可得到Nana能够抽取的最多牌数(取模后)。
请注意,这里假设n的范围不超过2000,并且给定的MOD值为1000000007。如果n的范围更大,可能需要使用其他方法进行优化。
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