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
TTT9

Sample Output
YES
//it's clear that player will not bid, then the blanker has only 2/13 winning percentage.

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抄袭、复制答案,以达到刷声望分或其他目的的行为,在CSDN问答是严格禁止的,一经发现立刻封号。是时候展现真正的技术了!
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Problem Description Shut the Box is a one-player game that begins with a set of N pieces labeled from 1 to N. All pieces are initially "unmarked" (in the picture at right, the unmarked pieces are those in an upward position). In the version we consider, a player is allowed up to T turns, with each turn defined by an independently chosen value V (typically determined by rolling one or more dice). During a turn, the player must designate a set of currently unmarked pieces whose numeric labels add precisely to V, and mark them. The game continues either until the player runs out of turns, or until a single turn when it becomes impossible to find a set of unmarked pieces summing to the designated value V (in which case it and all further turns are forfeited). The goal is to mark as many pieces as possible; marking all pieces is known as "shutting the box." Your goal is to determine the maximum number of pieces that can be marked by a fixed sequence of turns. As an example, consider a game with 6 pieces and the following sequence of turns: 10, 3, 4, 2. The best outcome for that sequence is to mark a total of four pieces. This can be achieved by using the value 10 to mark the pieces 1+4+5, and then using the value of 3 to mark piece 3. At that point, the game would end as there is no way to precisely use the turn with value 4 (the final turn of value 2 must be forfeited as well). An alternate strategy for achieving the same number of marked pieces would be to use the value 10 to mark four pieces 1+2+3+4, with the game ending on the turn with value 3. But there does not exist any way to mark five or more pieces with that sequence. Input Each game begins with a line containing two integers, N, T where 1 ≤ N ≤ 22 represents the number of pieces, and 1 ≤ T ≤ N represents the maximum number of turns that will be allowed. The following line contains T integers designating the sequence of turn values for the game; each such value V will satisify 1 ≤ V ≤ 22. You must read that entire sequence from the input, even though a particular game might end on an unsuccessful turn prior to the end of the sequence. The data set ends with a line containing 0 0. Output You should output a single line for each game, as shown below, reporting the ordinal for the game and the maximum number of pieces that can be marked during that game. Sample Input 6 4 10 3 4 2 6 5 10 2 4 5 3 10 10 1 1 3 4 5 6 7 8 9 10 22 22 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 Sample Output Game 1: 4 Game 2: 6 Game 3: 1 Game 4: 22

c++ 显示扑克牌牌面的系统函数

c++的系统函数中有没有显示扑克牌牌面的函数?像蜘蛛纸牌那样的,显示出整个牌面。求解答~

一个使用C语言模拟桌游策略纸牌的游戏,规则详见下面,怎么利用C语言解决

Problem Description Hearthstone is an online collectible card game from Blizzard Entertainment. Strategies and luck are the most important factors in this game. When you suffer a desperate situation and your only hope depends on the top of the card deck, and you draw the only card to solve this dilemma. We call this "Shen Chou Gou" in Chinese. Now you are asked to calculate the probability to become a "Shen Chou Gou" to kill your enemy in this turn. To simplify this problem, we assume that there are only two kinds of cards, and you don't need to consider the cost of the cards. -A-Card: If the card deck contains less than two cards, draw all the cards from the card deck; otherwise, draw two cards from the top of the card deck. -B-Card: Deal X damage to your enemy. Note that different B-Cards may have different X values. At the beginning, you have no cards in your hands. Your enemy has P Hit Points (HP). The card deck has N A-Cards and M B-Cards. The card deck has been shuffled randomly. At the beginning of your turn, you draw a card from the top of the card deck. You can use all the cards in your hands until you run out of it. Your task is to calculate the probability that you can win in this turn, i.e., can deal at least P damage to your enemy. Input The first line is the number of test cases T (T<=10). Then come three positive integers P (P<=1000), N and M (N+M<=20), representing the enemy’s HP, the number of A-Cards and the number of B-Cards in the card deck, respectively. Next line come M integers representing X (0<X<=1000) values for the B-Cards. Output For each test case, output the probability as a reduced fraction (i.e., the greatest common divisor of the numerator and denominator is 1). If the answer is zero (one), you should output 0/1 (1/1) instead. Sample Input 2 3 1 2 1 2 3 5 10 1 1 1 1 1 1 1 1 1 1 Sample Output 1/3 46/273

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[step-1][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://img-ask.csdn.net/upload/201504/27/1430134971_334457.png) ![图片说明](https://img-ask.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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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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《Oracle Java SE编程自学与面试指南》最佳学习路线图(2020最新版)

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