Mahjong Hu Pai algorithm series (1)

# Include <stdio. h>

Int Hu (INT Pai [38]);
Int remain (INT Pai [38]);

Int main ()
{
// ° N percent» ¸ ± å æ å â ã µ? ~~×öà ''' â {}° {{{~~~~~~~~~{{??
// ************** Pai [0], Pai [10], Pai [20], pai [30] ~~~~úö²» ó ~£%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
// Ö~~~=0 £ ¡
Int Pai [38] = {0,
1, 1, 0, 1, 1, 0, // Pai [1-9] *************************************
0,
0, 0, 0, 0, 0, 0
0,
0, 0, 0, 0, 0, 0, // Pai [21-29] Too many bytes faster
0,
0, 0, 0, 0, 0 // Pai [31-37] Then «ää î ÷ ± öð · ¢ ° × µäö öö ý
};
 
// Çademañeplus ± ø ~%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 201710000£ £ 000000000000000000000000²» ***********************************************
// If (qidui (PAI ))...
// If (shisanyao (PAI ))...
 
If (Hu (PAI ))
Printf ("Please have £;****************/N ");
Else
Printf ("° £ ");
 
Return 1;
}

// Adjust bandwidth ~è ý ç ¡± £%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
// °C °? à £;{%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Int Hu (INT Pai [38])
{
Static int Jiang = 0; // when «å æ ± ö ö£ ££ ¼ 'å æ í ° {° è ~~~~~~~~%%± öð ä¡ ° ¶ þ ¡±
 
If (! Remain (PAI) return 1; // µ ******************************************************* £.
 
For (INT I = 1 ;! Pai [I] & I <38; I ++ ); // zookeeper. zookeeper I have been chosen as the infrastructure Infrastructure infrastructure, and PAI [I] ç çoröö ö
 
Printf ("I = % d/N", I); // ¸ ú× ù å ï ¢
 
// 4 Gbit/s)
If (PAI [I] = 4) // %%%μ ± ç° %%%%%%%%%%%%%%%/4%%%%%%%%%%%%%
{
Pai [I] = 0;
If (Hu (PAI) return 1; // %% *******************************************************
Pai [I] = 4; // n has bought £;;%%%%%%4%%%% **************
}
 
// 3 Gbit/s)
If (PAI [I]> = 3) // %%%µ%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{
Pai [I]-= 3; // ********************
If (Hu (PAI) return 1; // %% *******************************************************
Pai [I] + = 3; // è ï û 3 3 then é ×é° then
}
 
// 2 Gbit/s)
If (! Jiang & PAI [I]> = 2) // â ç ö öç° ã» ó ð «£ £%%%%%µ%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%2%%%%%%%%%%%%%
{
Jiang = 1; // éööw.gov.cn
Pai [I]-= 2; // ~~=¥2 ~~~~~~
If (Hu (PAI) return 1; // %% *******************************************************
Pai [I] + = 2; // è ï û 2 then é ×é° then
Jiang = 0; // ç ³ ½ «å æ ± ö ö¾
}
 

If (I> 30) return 0; // ° ¶ «ää î ÷ ±?
 
// Please refer ¡
If (I % 10! = 8 & I % 10! = 9
Pai [I + 1] & PAI [I + 2]) // maid ~~~~~*************************************
{
Pai [I] --;
Pai [I + 1] --;
Pai [I + 2] --; // There are too many other than 1
If (Hu (PAI) return 1; // %% *******************************************************
Pai [I] ++;
Pai [I + 1] ++;
Pai [I + 2] ++;
}
 
¡
Return 0;
}

// **********************
Int remain (INT Pai [38])
{
Int sum = 0;
For (INT I = 1; I <38; I ++)
Sum + = Pai [I];
Return sum;
}

 

 

This algorithm is concise but problematic, and does not consider the absence of child. If this problem is not taken into consideration, when it is used in a game, a judgment error will occur. The above algorithm comes from a friend on the Internet. After my improvement, I have been able to determine the correct fake cards, as for the code, I will not post it. The people who visit my space basically do not reply, so I will not post some of my results.