题目:手里有五张牌,桌上有一堆牌(五张)。你能够弃掉手中的k张牌,然后从牌堆中取最上面的k个。
比較规则例如以下:(按优先级排序)
1.straight-flush:同花顺。牌面为T(10) - A,这里不论花色是否同样;
2.four-of-a-kind:四条,牌面有4个同样的值。
3.full-house:船牌,牌面有3个同样值,剩下2个也同样值。
4.flush:同花,五张牌的花色同样,不是同花顺;
5.straight:顺子。五张牌的值连续,A能够作为1也能够作为14;
6.three-of-a-kind:三条,牌面有3个同样的值;
7.two-pairs:两对。牌面有2个对子。
8.one-pair:一对,牌面有一个对子,即2个同值;
9.highest-card:大牌,没有以上牌型。
分析:搜索、枚举。枚举换牌数从0-5的每种情况,推断,取出优先值最高的就可以。
说明:读题是重点。
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio> #define min(x,y) ((x)<(y)? (x):(y)) using namespace std; char temp[5][3];
char card[10][3];
int maps[5][13]; char output[11][20] = {
"","straight-flush","four-of-a-kind",
"full-house","flush","straight","three-of-a-kind",
"two-pairs","one-pair","highest-card",""}; int value( char ch )
{
if ( ch == 'T' ) return 9;
if ( ch == 'J' ) return 10;
if ( ch == 'Q' ) return 11;
if ( ch == 'K' ) return 12;
if ( ch == 'A' ) return 0;
return ch - '1';
} int color( char ch )
{
if ( ch == 'S' ) return 0;
if ( ch == 'H' ) return 1;
if ( ch == 'D' ) return 2;
if ( ch == 'C' ) return 3;
} int tests()
{
//royal-flush | straight-flush
for ( int i = 0 ; i < 5 ; ++ i )
if ( maps[i][0]&maps[i][9]&maps[i][10]&maps[i][11]&maps[i][12] )
return 1;
//four-of-a-kind
for ( int i = 0 ; i < 13 ; ++ i )
if ( maps[4][i] == 4 ) return 2;
//full-house
int three = 0,two = 0;
for ( int i = 0 ; i < 13 ; ++ i ) {
if ( maps[4][i] == 2 ) two ++;
if ( maps[4][i] == 3 ) three ++;
}
if ( two && three ) return 3;
//flush
for ( int i = 0 ; i < 4 ; ++ i ) {
int count = 0;
for ( int j = 0 ; j < 13 ; ++ j )
count += maps[i][j];
if ( count >= 5 ) return 4;
}
//straight
for ( int i = 0 ; i < 10 ; ++ i )
if ( maps[4][i]&maps[4][i+1]&maps[4][i+2]&maps[4][i+3]&maps[4][(i+4)%13] )
return 5;
//three-of-a-kind
if ( three ) return 6;
//two-pairs
if ( two > 1 ) return 7;
//one-pair
if ( two ) return 8;
return 9;
} void change()
{
for ( int i = 0 ; i < 5 ; ++ i )
strcpy(temp[i],card[i+5]);
memset( maps, 0, sizeof(maps) );
for ( int i = 0 ; i < 5 ; ++ i ) {
maps[color(temp[i][1])][value(temp[i][0])] = 1;
maps[4][value(temp[i][0])] ++;
}
int Min = tests(); for ( int k = 1 ; k <= 5 ; ++ k ) {
int xx,yy,comb = (1<<k)-1;
while ( comb < 32 ) {
// 计算当前状态相应的集合
int j = 0,count = 0,move = 5;
do {
if ( (1<<j)&comb )
strcpy(temp[count ++],card[j]);
j ++;
}while ( j < 5 ); while ( count < 5 )
strcpy(temp[count ++],card[move ++]); memset( maps, 0, sizeof(maps) );
for ( int i = 0 ; i < 5 ; ++ i ) {
maps[color(temp[i][1])][value(temp[i][0])] = 1;
maps[4][value(temp[i][0])] ++;
} Min = min(Min,tests());
// 位运算计算下一组合
xx = comb&-comb,yy = comb+xx;
comb = ((comb&~yy)/xx>>1)|yy;
}
}
printf("%s\n",output[Min]);
} int main()
{
while ( ~scanf("%s",card[0]) ) {
for ( int i = 1 ; i < 10 ; ++ i )
scanf("%s",card[i]); printf("Hand: ");
for ( int i = 0 ; i < 5 ; ++ i )
printf("%s ",card[i]);
printf("Deck: ");
for ( int i = 5 ; i < 10 ; ++ i )
printf("%s ",card[i]);
printf("Best hand: ");
change();
}
return 0;
}