131 - The Psychic Poker Player（直接枚举）
题目：131 - The Psychic Poker Player
题目大意：刚开始我一点都不懂题目要我们做什么，后来看了别人的题解，发现原来是打某种类型的纸牌，手上有5张牌，桌上也五张牌，这个人可艺从手中丢弃某些牌，从轴上摸去相等数量的牌，桌上的牌是有顺序的，然后问手上最好的牌是什么。
解题思路：直接枚举丢牌和摸牌的情况，一共有2的5次方种，这里我用了位运算，用一个数来代替舍弃牌的情况，（这个数的二进制数某个位置上有1，代表舍弃这个位置上的数），将这个数与1,10,100,1000,10000相与，就可以知道哪个位置上有1了，这样就可以知道舍弃哪个位置上的牌。剩下的就是得判断手上的是什么牌，细心就可以过得，题目给的代码能过一般就没问题了。
注意：每次牌都还得回到原来初始的状态。还有里面有字母TK什么的需要处理成数字比较方便。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N = 10;
const char ans[N][2*N] = {"straight-flush", "four-of-a-kind", "full-house", "flush", "straight", "three-of-a-kind", "two-pairs", "one-pair", "highest-card"};
int choice;
int str[5], color[5];
char s[10][4];
int is_kind() {
int num[15], i;
memset(num, 0, sizeof(num));
for(i = 0; i < 5; i++)
num[str[i]]++;
sort(num, num + 15);
int count = 0;
for(i = 14; i >= 0; i--) {
if(num[i] == 4)
return 1;
else if(num[i] == 3) {
for(int j = i - 1; j >= 0; j--)
if(num[j] == 2)
return 2;
return 5;
}
else if(num[i] == 2)
count++;
}
if(count == 2)
return 6;
else if(count == 1)
return 7;
else return -1;
}
void manage() {
int i;
for(i = 0; i < 5; i++) {
if(s[i][0] == 'A')
str[i] = 1;
else if(s[i][0] == 'T')
str[i] = 10;
else if(s[i][0] == 'J')
str[i] = 11;
else if(s[i][0] == 'Q')
str[i] = 12;
else if(s[i][0] == 'K')
str[i] = 13;
else str[i] = s[i][0] - '0';
}
for(i = 0; i < 5; i++) {
if(s[i][1] == 'C')
color[i] = 1;
else if(s[i][1] == 'D')
color[i] = 2;
else if(s[i][1] == 'H')
color[i] = 3;
else
color[i] = 4;
}
}
int is_Flush() {
int count = 0;
for(int i = 1; i < 5; i++)
if(color[i] == color[0])
count++;
if(count == 4)
return 1;
return 0;
}
int is_Straight() {
int count = 0, i;
int s1[5];
for(i = 0; i < 5; i++)
s1[i] = str[i];
sort(s1, s1 + 5);
for(i = 1; i < 5; i++){
if(s1[i] == s1[i - 1] + 1 )
count++;
}
if(count == 4)
return 1;
count = 0;
if(s1[0] == 1 && s1[1] == 10) {
for(i = 2; i < 5; i++)
if(s1[i] == s1[i - 1] + 1)
count++;
if(count == 3)
return 1;
}
return 0;
}
int find() {
int c = is_kind();
if( c == -1) {
if(is_Straight() ) {
if(is_Flush())
c = 0;
else
c = 4;
}
else {
if(is_Flush())
c = 3;
else
c = 8;
}
}
else {
if(is_Flush() && c > 3)
c = 3;
}
return c;
}
int main() {
while(1) {
int i, j;
for(i = 0; i < 10; i++) {
if(!i) {
if(scanf("%s", s[0]) != 1)
break;
}
else
scanf("%s", s[i]);
}
if(i == 0)
break;
choice = 10;
for(i = 0; i < (1 << 5); i++) {
manage();
int n = 0, d[5];
memset(d, 0, sizeof(d));
for(j = 0; j < 5; j++)
if(i & (1 << j)) {
a n++;
d[j]++;
}
int k = 5, v;
for(v = 0; v < 5; v++)
if(d[v]) {
if(s[k][0] == 'A')
str[v] = 1;
else if(s[k][0] == 'T')
str[v] = 10;
else if(s[k][0] == 'J')
str[v] = 11;
else if(s[k][0] == 'Q')
str[v] = 12;
else if(s[k][0] == 'K')
str[v] = 13;
else str[v] = s[k][0] - '0';
if(s[k][1] == 'C')
color[v] = 1;
else if(s[k][1] == 'D')
color[v] = 2;
else if(s[k][1] == 'H')
color[v] = 3;
else
color[v] = 4;
k++;
}
int m = find();
if(m < choice)
choice = m;
}
printf("Hand: ");
for(i = 0; i < 5; i++)
printf("%s ", s[i]);
printf("Deck: ");
for(i = 5; i < 10; i++)
printf("%s ", s[i]);
printf("Best hand: %s\n", ans[choice]);
}
return 0;
}