topic
Dongdong has A × B playing cards. Each playing card has a size (integer, a, range 0 to A-1) and a suit (integer, b), range 0 to B-1.
Playing cards are different, also It is unique, that is to say, no two cards are the same size and suit.
"One hand" means that you have 5 different cards in your hand. These 5 cards are not in the order of who is in front and who is behind. They can form a card type. We have defined 9 card types. The following are the rules of 9 card types. We use "low serial number priority" to match the card types, that is, the "first hand" meets the first from the top. A card type rule is its "card number" (an integer, from 1 to 9):
Straight flush: Rule 5 and rule 4 are both met.
Bomb: 5 cards of which 4 cards are the same size.
Three belts two: Among the 5 cards, 3 of them are the same size, and the other 2
are of the same
size . Straight: 5 cards are of the same suit. Straight: 5 cards of the same size as x, x + 1, x + 2, x + 3, x + 4
three: 5 cards which is equal to the size of three cards.
two pairs: 5 cards which is equal to the size of two cards, Further card 2 is equal to three card size.
One pair: 5 cards which is equal to the size of two cards.
To afford: the hand does not satisfy any of a card type.
Now, from Eastern A × 2 cards were taken from the B playing cards! They are (a1, b1) and (a2, b2). (Where a is the size and b is the suit)
Now we will randomly take out the remaining cards 3 cards! Form a hand card !!!
In fact, in addition to playing codes, Dongdong is still a magician in his spare time. Now he wants to predict the possibility of his future, that is, the possibility of the "first hand" he will get. , Belonging to 1 to 9) "to indicate the hand type of this hand, then he has 9 possibilities in the future, but the number of each possible scheme is different.
Input
Line 1 contains integers A and B (5 ≤ A ≤ 25, 1 ≤ B ≤ 4).
Line 2 contains integers a1, b1, a2, b2 (0 ≤ a1, a2 ≤ A-1, 0 ≤ b1 , b2 ≤ B-1, (a1, b1) ≠ (a2, b2)).
Output
A line is output. There are 9 integers in this line, and each integer represents the number of schemes of 9 card types (in order of card number from small to large).
Sample Input
5 2
1 0 3 1
Sample Ouput
0 0 0 0 8 0 12 36 0
Ideas
The type of cards is 4 * 25 = 100 at most, and the amount of data is small. Choose 3 cards in a direct three-layer loop with a complexity of O (n 3 ), which is acceptable.
Add all the card types (except the two given) to the array of structure card. The three-layer loop of the array selects three cards and judges the card type of a hand.
In order to compare the size of the cards, makeOrdered cards. After the 5 cards are ordered, you only need to compare the size of the cards in the special position to determine the card type. For example, ifNumber 0 and Number 3Of the same size orNumber 1 and Number 4Is the same size asbomb。
Code
#include <cstdio>
#include <algorithm>
using namespace std;
struct card{
int suit,val;
card& operator=(card c){suit=c.suit;val=c.val;return *this;}
bool operator==(card c){return suit==c.suit&&val==c.val;}
bool operator<(card c)const{return val!=c.val?val<c.val:suit<c.suit;}
}c[105],tmp[6];
int sum[10];
int A,B;
void judge(){
bool flags=true,flagv=true;
for(int i=0;i<4;i++){
if(tmp[i+1].val-tmp[i].val!=1)
flagv=false;
if(tmp[i+1].suit!=tmp[i].suit)
flags=false;
}
if(flags&&flagv)//同花顺
sum[1]++;
else if(flags)//同花
sum[4]++;
else if(flagv)//顺子
sum[5]++;
else{
if((tmp[0].val==tmp[3].val)||(tmp[1].val==tmp[4].val))//炸弹
sum[2]++;
else if((tmp[0].val==tmp[1].val&&tmp[2].val==tmp[4].val)||(tmp[0].val==tmp[2].val&&tmp[3].val==tmp[4].val))//三带二
sum[3]++;
else if(tmp[0].val==tmp[2].val||tmp[2].val==tmp[4].val||tmp[1].val==tmp[3].val)//三条
sum[6]++;
else if(tmp[0].val==tmp[1].val&&tmp[2].val==tmp[3].val)//两对
sum[7]++;
else if(tmp[0].val==tmp[1].val&&tmp[3].val==tmp[4].val)//两对
sum[7]++;
else if(tmp[1].val==tmp[2].val&&tmp[3].val==tmp[4].val)//两对
sum[7]++;
else{
bool flag=false;
for(int i=0;i<4;i++){
if(tmp[i+1].val==tmp[i].val)
flag=true;
}
if(flag)
sum[8]++;
else sum[9]++;
}
}
}
int main(){
memset(sum, 0, sizeof(sum));
scanf("%d%d",&A,&B);
int a1,b1,a2,b2;
scanf("%d%d%d%d",&a1,&b1,&a2,&b2);
int k=0;
for(int i=0;i<A;i++){
for(int j=0;j<B;j++){
if(!(i==a1&&j==b1)&&!(i==a2&&j==b2)){
c[k].val=i;c[k].suit=j;
k++;
}
}
}
for(int i=0;i<k;i++){
for(int j=i+1;j<k;j++){
for(int m=j+1;m<k;m++){
tmp[0].val=a1;tmp[0].suit=b1;
tmp[1].val=a2;tmp[1].suit=b2;
tmp[2]=c[i];
tmp[3]=c[j];
tmp[4]=c[m];
sort(tmp,tmp+5);
judge();
}
}
}
for(int i=1;i<=9;i++)
printf("%d ",sum[i]);
return 0;
}
to sum up
At first I thought about recursive enumeration, but obviously I won't. .
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