The Psychic Poker Player

I have five cards on my hand and five cards on the stack. Now I want to discard the n (0 <= n <= 5) cards on my hand, take n cards from the heap. If you have a special function, you can know what the five cards on the heap are from top to bottom. You must obtain the largest card at the end, A card consists of numbers and colors.

Train of Thought: I will give you one by one each kind of case and then judge (children who do not know how to play cards cannot afford to hurt this question)

Code:
# Include <stdio. h>
# Include <string. h>
# Include <stdlib. h>
Struct node
{
Char str [3];
} E [12], ee [12], ans [12];
Int p = 0, max = 0;
Char str [] [19] = {"highest-card", "one-pair", "two-pairs", "three-of-a-kind ",
"Straight", "flush", "full-house", "four-of-a-kind", "straight-flush "};
Char str1 [19] = "123456789:; <=", str2 [10] = "CDHS ";
Int cmp (void const * a, void const * B)
{
Return (* (struct node *) a). str [0]-(* (struct node *) B). str [0];
}
Void change (int p)
{
If (ee [p]. str [0] = 'A') {ee [p]. str [0] = '1'; return ;}
If (ee [p]. str [0] = 'T') {ee [p]. str [0] = ':'; return ;}
If (ee [p]. str [0] = 'J') {ee [p]. str [0] = '; return ;}
If (ee [p]. str [0] = 'q') {ee [p]. str [0] = '<'; return ;}
If (ee [p]. str [0] = 'k') {ee [p]. str [0] = '; return ;}
Return;
}
Int judge ()
{
Int I = 0, p = 0, num1 [5], num2 [5];
Char ans1 [10], ans2 [10];
Qsort (ans, 5, sizeof (ans [0]), cmp); // The characters in ans can be regarded as numbers from 1 to 13.
For (I = 0; I <5; I ++)
{
Num1 [I] = num2 [I] = 1;
}
For (I = 0; I <5; I ++)
{
Ans1 [I] = ans [I]. str [0]; // obtain the nominal value
Ans2 [I] = ans [I]. str [1]; // color
If (I)
{
If (ans1 [I] = ans1 [I-1]) // record the number of times
Num1 [I] + = num1 [I-1];
If (ans2 [I] = ans2 [I-1])
Num2 [I] + = num2 [I-1];
}
}
P = ans1 [0]-'1 ';
If (p <10 & strncmp (ans1, str1 + p, 5) = 0 & num2 [4] = 5)
Return 8;
If (ans1 [0] = '1' & ans1 [1] = ':' & ans1 [2] = '; '& ans1 [3] =' <'& ans1 [4] =' & num2 [4] = 5)
Return 8;
If (num1 [3] = 4 | num1 [4] = 4)
Return 7;
If (num1 [2] = 3 & num1 [4] = 2) | (num1 [1] = 2 & num1 [4] = 3 ))
Return 6;
If (num2 [4] = 5)
Return 5;
If (p <10 & strncmp (ans1, str1 + p, 5) = 0)
Return 4;
If (ans1 [0] = '1' & ans1 [1] = ':' & ans1 [2] = '; '& ans1 [3] =' <'& ans1 [4] = ')
Return 4;
If (num1 [4] = 3 & num1 [0] = 1 & num1 [1] = 1) | (num1 [3] = 3 & num1 [0] = 1 & num1 [4] = 1) | (num1 [2] = 3 & num1 [4] = 1 ))
Return 3;
If (num1 [1] = 2 & num1 [3] = 2 & num1 [4] = 1) | (num1 [2] = 2 & num1 [4] = 2) | (num1 [1] = 2 & num1 [4] = 2 ))
Return 2;
If (num1 [2] = 2) | (num1 [3] = 2) | (num1 [4] = 2) | (num1 [1] = 2 ))
Return 1;
Return 0;
}
Void create (int cu, int p) // enumerate each case
{
Int I = 0, j = 0, k = 0;
If (cu = 6)
{
For (j = p, I = 5; I <= 9-p; I ++)
{
Strcpy (ee [j ++]. str, e [I]. str );
Change (J-1 );
}
For (I = 0; I <5; I ++)
Strcpy (ans [I]. str, ee [I]. str );
K = judge ();
Max = max> k? Max: k;
Return;
}
For (I = 0; I <2; I ++)
{
If (I)
{
Strcpy (ee [p]. str, e [cu-1]. str );
Change (p );
Create (cu + 1, p + 1 );
}
Else
Create (cu + 1, p );
}
}
Int main ()
{
Int I = 0;
While (scanf ("% s", e [0]. str )! = EOF)
{
Printf ("Hand :");
Printf ("% s", e [0]. str );
For (I = 1; I <10; I ++)
{
Scanf ("% s", e [I]. str );
If (I = 5)
Printf ("Deck :");
Printf ("% s", e [I]. str );
}
Printf ("Best hand :");
Max = p = 0;
Create (1, p );
Printf ("% s \ n", str [max]);
} Www.2cto.com
Return 0;
}

Author: ulquiorra0cifer