Cow and Cow game algorithm

<?phpnamespace frontend\business;classniuniugamehelper{/** * @param $card * @return int Results-1 per cow, 0 kn, 1-9 kn to ox nine*/     Public Static functionJudgecowcow ($card)    {        if(!Is_array($card) ||Count($card)!== 5)        {            return-1; }        $cow=-1; //calculate the total value of 5 cards, cow calculate the number of cows.         $cardall= 0; $n= 0;//Store 10, J, Q, K number.          for($i= 0;$i<4;$i++)//sort from big to small for 5 cards.         {             for($j=$i+1;$j<5;$j++)                    if($card[$i] <$card[$j])                    {                        $a=$card[$i]; $card[$i] =$card[$j]; $card[$j]=$a; }        }         for($i= 0;$i<5;$i++)        {            if($card[$i] >= 10)            {                $n++; $card[$i] = 10; }            $cardall+=$card[$i]; }        //10 10 6 5 3 NIU NIU 5 card sum must be a multiple of 10, five card sum is 10 of the multiples may not be bull cattle, the following code is not established//if (cardall%10 = = 0)//{//print (0);        Niuniu//exit (0); //}        Switch($n)        {             Case0://5 cards do not have a 10, J, Q, K.             {                 for($i= 0;$i<4;$i++)                {                     for($j=$i+ 1;$j<5;$j++)                        if(($cardall-$card[$i]-$card[$j])%10==0)                        {                            $cow=($card[$i] +$card[$j])%10; return $cow; }                }                 Break; }             Case1://5 cards have a 10, J, Q, K.             {                //first of all to determine whether there are cattle and cattle, can not judge the remaining four added to 10 times times the number of cattle, such as Q 8 5 4 3//can only determine whether the two is a multiple of 10, if it is to determine whether the surplus is a multiple of 10, the limited judgment of the cow and cattle, and then to determine whether three is a multiple Cattle, otherwise no cattle                 for($i= 1;$i< 4;$i++)                {                     for($j=$i+1;$j< 5;$j++)                    {                        if(($card[$i] +$card[$j])% 10 = = 0)                        {                            $cow=($cardall-$card[0]) %10; return $cow; }                    }                }                //determine if there are any cows                 for($i= 1;$i<5;$i++)//There are four cards left and three plus equals ten.                {                    if(($cardall-$card[0]-$card[$i])%10==0)                    {                        $cow=($cardall-$card[0]) %10;  Break; }                }                 Break; }             Case2://5 cards have two 10, J, Q, K. Three is a cow, you have a problem, you should prioritize the output            {                if(($cardall-$card[0]-$card[1]) %10 = = 0)//Priority cow and ox output such as J Q 2 3 5; Check if the remainder is bull or bull, otherwise the algorithm has a loophole.                {                    $cow= 0; }                Else                {                    //Ten 6 5 3 n=2 i=3 j=4 cardall =                     for($i=$n;$i<4;$i++)//The remaining three (four) cards have two and add up equal to 10.                     {                         for($j=$i+1;$j<5;$j++)                        {                            if(($card[$i]+$card[$j]) ==10)                            {                                $temp=$cardall;  for($k= 0;$k<$n;$k++)//the total value minus 10, J, Q, K cards.                                     $temp-=$card[$k];// -                                $cow=$temp%10//8                            }                        }                        /*Print_r (' Cardall: '. $cardall);                        Print_r ("<br/>");                        Print_r (' cow '. $cow); Print_r ("<br/>");*/                    }                }                 Break; }             Case3://5 cards have three 10, J, Q, K.              Case4://5 cards have four 10, J, Q, K.              Case5://5 cards in five are 10, J, Q, K.             {                 for($i= 0;$i<$n;$i++)//the total value minus 10, J, Q, K cards.                 {                    $cardall-=$card[$i]; }                $cow=$cardall%10;  Break; }        }        return $cow; }} 

Cow and Cow game algorithm