 扑克牌获胜概率的计算问题，用C语言来实现，具体怎么做的？

Problem Description
21 point,also named Black Jack, originated in France,has spread around the world.21 point, a gambling game played by using poker,is also the only one which is able to win banker by using probability calculation.
 Encyclopedia from good search
We define blackjack rules are as follows, which is different from the original rules slightly.
cards are as follows:
A 2 3 4 5 6 7 8 9 10 J Q K
A is as 1 point.
JQK are all as 10 points.
We assume that the casino prepared a lot of cards,that is, you can assume that the probability of getting each card is the same.
There are two players who were banker and Player.
They get two cards at first and can see the card each other.
Player operates first.
Every turn, he can bid or stop bidding.
If he bid, he can take a card from the deck.
Once the the points are over 21,he will lose at once,which is called "busting", otherwise he will bid until stopping bidding and turn to banker.
the rule of banker is the same as the player's.
If there is no "busting", the one who have higher points win.
If they have the same points,they get the tie.
Here is the task,we give you the cards that both people have gotten.
please judge whether the Player have more than 50% winning percentage,
if yes, output "YES", otherwise output,"NO".Oh, yes, everyone is very smart.
Input
The first line of the input file contains an integer Test(Test<=100000), which indicates the number of test cases.For each test case, there is a string with four characters. the first and the second char indicate the card of the player.and the others indicate the banker's.(we use ‘T’ instead of '10')
Output
For each case of data output "YES" or "NO", it indicates whether the Player have more than 50% winning percentage.Sample Input
1
TTT9Sample Output
YES
//it's clear that player will not bid, then the blanker has only 2/13 winning percentage.
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 用C语言解答，一个21点的扑克牌的算法，看是否可以算出来？
Problem Description 21 point,also named Black Jack, originated in France,has spread around the world. 21 point, a gambling game played by using poker,is also the only one which is able to win banker by using probability calculation.  Encyclopedia from good search We define blackjack rules are as follows, which is different from the original rules slightly. cards are as follows: A 2 3 4 5 6 7 8 9 10 J Q K A is as 1 point. JQK are all as 10 points. We assume that the casino prepared a lot of cards,that is, you can assume that the probability of getting each card is the same. There are two players who were banker and Player. They get two cards at first and can see the card each other. Player operates first. Every turn, he can bid or stop bidding. If he bid, he can take a card from the deck. Once the the points are over 21,he will lose at once,which is called "busting", otherwise he will bid until stopping bidding and turn to banker. the rule of banker is the same as the player's. If there is no "busting", the one who have higher points win. If they have the same points,they get the tie. Here is the task,we give you the cards that both people have gotten. please judge whether the Player have more than 50% winning percentage, if yes, output "YES", otherwise output,"NO". Oh, yes, everyone is very smart. Input The first line of the input file contains an integer Test(Test<=100000), which indicates the number of test cases. For each test case, there is a string with four characters. the first and the second char indicate the card of the player.and the others indicate the banker's.(we use ‘T’ instead of '10') Output For each case of data output "YES" or "NO", it indicates whether the Player have more than 50% winning percentage. Sample Input 1 TTT9 Sample Output YES //it's clear that player will not bid, then the blanker has only 2/13 winning percentage.
 采用深度优先搜索进行扑克牌的排序
``` #include<iostream> using namespace std; int count=0; int book[5]; char card[5][2]={'2','C','A','D','A','C','J','C','J','H'}; char a[5][2]; void dfs(int step){ if(step==5){ count++; return ; } for(int i=0;i<5;i++){ if(/*a[step][0]!=a[step1][0]&&*/book[i]==0){//判断条件该牌未被标记且与上一张的牌号不相同，但我不知道应该如何写出这条判断语句，如果我将其注释掉，则输出120，是一个全排列。如果不注释，则输出为0. a[step][0]=card[i][0]; book[i]=1;//将用过的牌标记 dfs(step+1); book[i]=0;//取消标记 } } return ; } int main(){ dfs(0); cout<<count<<endl; return 0; } ``` ![图片说明](https://imgask.csdn.net/upload/201504/27/1430134971_334457.png) ![图片说明](https://imgask.csdn.net/upload/201504/27/1430134985_479660.png) 我未按照题目要求，只是给出了特定的扑克牌进行算法可行性的验证，如第三组数据所示，该程序应输出48. 现在的问题是不知如何进行条件的判断。悉心向各位请教
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 扑克牌的牌技问题，怎么利用C语言的程序的设计方法对此问题加以解决的
Problem Description The magician shuffles a small pack of cards, holds it face down and performs the following procedure: 1.The top card is moved to the bottom of the pack. The new top card is dealt face up onto the table. It is the Ace of Spades. 2.Two cards are moved one at a time from the top to the bottom. The next card is dealt face up onto the table. It is the Two of Spades. 3.Three cards are moved one at a time… 4.This goes on until the nth and last card turns out to be the n of Spades. This impressive trick works if the magician knows how to arrange the cards beforehand (and knows how to give a false shuffle). Your program has to determine the initial order of the cards for a given number of cards, 1 ≤ n ≤ 13. Input On the first line of the input is a single positive integer, telling the number of test cases to follow. Each case consists of one line containing the integer n. Output For each test case, output a line with the correct permutation of the values 1 to n, space separated. The first number showing the top card of the pack, etc… Sample Input 2 4 5 Sample Output 2 1 4 3 3 1 4 5 2
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