UVa 131 The Psychic Poker Player

题目大意：手上有5张牌，堆上有5张牌，现在要求舍弃手上的n（0<=n<=5）牌，从堆的面上拿n张牌，假设了你有特异功能，可以知道堆上的5张牌从上到下是是什么，要求最后得到的牌要面值最大，一张牌由数字与花色组成。

思路：就是一一枚举出每种请况，然后判断（不懂打牌的孩子这道题伤不起啊）

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);//ans中的字符可以看成1到13的数字
    for(i = 0; i<5; i++)
    {
        num1[i] = num2[i] = 1;
    }
    for(i = 0; i<5; i++)
    {
        ans1[i] = ans[i].str[0];//取面值
        ans2[i] = ans[i].str[1];//取花色
        if(i)
        {
            if(ans1[i] == ans1[i-1])//记录出现次数
                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)//枚举每种情况
{
    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);
        }
&n

补充：软件开发 , C语言 ,