求高手帮忙写一个c语言发牌程序

#include <time.h>

#include <stdio.h>

#include <conio.h>

#include <stdlib.h>

#define PLAYER 4//玩家人数

#define NUM 13//玩家拿牌数

#define SIZE 52//所有牌数

//声明函数

void PokerRand(int *pokerRand)

void Palyer(int *pokerRand)

void Process(int *countA, int *countB, int *countC, int *countD)

void Output(int *poker, int *countA, int *countB, int *countC, int *countD)

struct PokerGame

{

int A[NUM]//记录玩家手中的黑桃牌

int B[NUM]//记录玩家手中的红桃牌

int C[NUM]//记录玩家手中的梅花牌

int D[NUM]//记录玩家手中的方片牌

int manNum[NUM]//记录玩家手里所有的牌

}man[PLAYER]

//随机产生52张牌

void PokerRand(int *pokerRand)

{

int i, j

srand((unsigned)time(NULL))

for (i=0i<SIZEi++)

{

MARK:pokerRand[i] = rand()%52

for (j=0j<ij++)

{

if (pokerRand[i] == pokerRand[j])

{

goto MARK

}

}

}

}

//给4个玩家发牌

void Palyer(int *pokerRand)

{

int i, j

int count = 0

for (j=0j<NUMj++)

{

for (i=0i<PLAYERi++)//轮流发牌

{

man[i].manNum[j] = pokerRand[count++]

}

}

}

//统计玩家手中的牌

void Process(int *countA, int *countB, int *countC, int *countD)

{

int i, j

for (i=0i<PLAYERi++)

{

countA[i] = 0

countB[i] = 0

countC[i] = 0

countD[i] = 0

for (j=0j<NUMj++)//统计四个玩家手中的牌

{

if ((man[i].manNum[j] >= 0) &&(man[i].manNum[j] <13))//黑桃

{

man[i].A[ countA[i]++ ] = man[i].manNum[j]

}

else if (man[i].manNum[j] <26)//红桃

{

man[i].B[ countB[i]++ ] = man[i].manNum[j]

}

else if (man[i].manNum[j] <39)//梅花

{

man[i].C[ countC[i]++ ] = man[i].manNum[j]

}

else//方片

{

man[i].D[ countD[i]++ ] = man[i].manNum[j]

}

}

}

}

//输出

void Output(int *poker, int *countA, int *countB, int *countC, int *countD)

{

int i, j

printf("扑克牌自动发牌 %c(黑) %c(红) %c(梅) %c(方):\n", 6, 3, 5, 4)

for (i=0i<PLAYERi++)

{

printf("\n第%d人 :\n", i+1)//开始输出第i个玩家

printf("%c:\t", 6)//输出第i个玩家的黑桃牌

for (j=0j<countA[i]j++)

{

if (poker[ man[i].A[j] ] == 10)//假如等于10，以%d格式输出

{

printf("%4d", poker[ man[i].A[j] ])

}

else//否则以%c格式输出

{

printf("%4c", poker[ man[i].A[j] ])

}

}

printf("\n")

printf("%c:\t", 3)//输出第i个玩家的红桃牌

for (j=0j<countB[i]j++)

{

if (poker[ man[i].B[j] ] == 10)

{

printf("%4d", poker[ man[i].B[j] ])

}

else

{

printf("%4c", poker[ man[i].B[j] ])

}

}

printf("\n")

printf("%c:\t", 5)//输出第i个玩家的梅花牌

for (j=0j<countC[i]j++)

{

if (poker[ man[i].C[j] ] == 10)

{

printf("%4d", poker[ man[i].C[j] ])

}

else

{

printf("%4c", poker[ man[i].C[j] ])

}

}

printf("\n")

printf("%c:\t", 4)//输出第i个玩家的方片牌

for (j=0j<countD[i]j++)

{

if (poker[ man[i].D[j] ] == 10)

{

printf("%4d", poker[ man[i].D[j] ])

}

else

{

printf("%4c", poker[ man[i].D[j] ])

}

}

printf("\n")

}

}

void main(void)

{

int countA[PLAYER] = { 0 }//记录4个玩家持黑桃牌数

int countB[PLAYER] = { 0 }//记录4个玩家持红桃牌数

int countC[PLAYER] = { 0 }//记录4个玩家持梅花牌数

int countD[PLAYER] = { 0 }//记录4个玩家持方片牌数

int pokerRand[SIZE] = { 0 }//存放随机产生52张牌

int poker[SIZE] = {65, 50, 51, 52, 53, 54, 55, 56, 57, 10, 74, 81, 75,

65, 50, 51, 52, 53, 54, 55, 56, 57, 10, 74, 81, 75,

65, 50, 51, 52, 53, 54, 55, 56, 57, 10, 74, 81, 75,

65, 50, 51, 52, 53, 54, 55, 56, 57, 10, 74, 81, 75,}

PokerRand(pokerRand)//洗牌

Palyer(pokerRand)//发牌

Process(countA, countB, countC, countD)//整牌

Output(poker, countA, countB, countC, countD)//亮牌

printf("\n\n\n")

system("pause")

}

这是以随机的形式发牌的。希望能帮到你。

int main(){int a[60] int b[60] int w = 0 int wi,wj,wii int i,j,index,x=0,m for(i=0i<54i++)a[i]=i+1 srand((unsigned)time(0)) for(i=0i<54){index=rand()%54 if(a[index]!=0){b[i]=a[index] //printf("%d ",b[i]) a[index]=0 ++i }}printf("\n说明：\n") printf("*******************\n") printf("**红桃 %c ",3) printf("**\n") printf("**方块 %c ",4) printf("**\n") printf("**梅花 %c ",5) printf("**\n") printf("**黑桃 %c ",6) printf("**\n") printf("**大王 A **\n**小王 a **\n") printf("*******************\n") printf("\n请输入打牌人数：\n") scanf("%d",&w) wi = 54/w wj = 54%w //printf("%d,%d",wi,wj) for (j = 0j <wj++){m = 0 printf("\n\n第%d个人的牌\n",j+1) wj -- if (wj>=0)wii = wi + 1 elsewii = wi//红桃printf("红桃 %c ",3) printf(":") for(i = j*wiii<(j+1)*wiii++){if (b[i]<=13&&b[i]>0){m++ if (b[i]==11)printf("J ") else if (b[i]==12)printf("Q ") else if (b[i]==13)printf("K ") elseprintf("%d ",b[i]) }} //方块printf("\n方块 %c ",4) printf(":") for(i = j*wiii<(j+1)*wiii++){if (b[i]<=26&&b[i]>13){m++ b[i] = b[i]-13 if (b[i]==11)printf("J ") else if (b[i]==12)printf("Q ") else if (b[i]==13)printf("K ") elseprintf("%d ",b[i]) }} //梅花printf("\n梅花 %c ",5) printf(":") for(i = j*wiii<(j+1)*wiii++){ if (b[i]<=39&&b[i]>26){m++ b[i] = b[i]-26 if (b[i]==11)printf("J ") else if (b[i]==12)printf("Q ") else if (b[i]==13)printf("K ") elseprintf("%d ",b[i]) }}//黑桃printf("\n黑桃 %c ",6) printf(":") for(i = j*wiii<(j+1)*wiii++){if (b[i]<=52&&b[i]>39){m++ b[i] = b[i]-39 if (b[i]==11)printf("J ") else if (b[i]==12)printf("Q ") else if (b[i]==13)printf("K ") elseprintf("%d ",b[i]) }}//大小王printf("\n大小王 A or a :") for(i = j*wiii<=(j+1)*wiii++){if (b[i]==53){m++ printf("A ") }if (b[i]==54){m++ printf("a ") }}printf("\n总共：%d张牌！！！",m) } }

其实发牌问题里有个随机的问题，就是模拟洗牌的问题，我来尝试下这个程序吧(每个花色的代码分别是：红桃 \x3,方块\x4,梅花\x5,黑桃\x6

#include <stdio.h>

#include <time.h>

char hua_se[4]={'\x3','\x4','\x5','\x6'} //定义一个数组来存放花色

char *dight[14]={"Ace","two","three","four","five","six","seven","eight","nine","ten","jack","queen","king"}//定义一个指针型的数组来存放数字

int fa_pai[4][13]={0}//定义一个二维数组来发牌

void move(int fa_pai[4][13])

void deal(int fa_pai[4][13],char huase[4],char *dight[4][13])

int main(void){srand(time(NULL))

move(fa_pai)

deal(hua_se,dight,fa_pai)

printf("你是否想结束发牌？Y/N")

getch()return 0}void move(int fa_pai[4][13]){int r,card,row,column

for(card=1card<=52card++)

{

r=rand()

row=r%4

r=rand()

column=r%13

while(fa_pai[4][13]!=0)

{

r=rand()

row=r%4

r=rand()

column=r%13

}

fa_pai[row][column]=card

}

}

void deal(int fapai[4][13],char hua_se[3],char *dight[13])

{

char cint card,row,column

for(card=1card<=52card++)

{

for(row=0row<=3row++)

{

for(column=0column<=12column++)

{

if(fapai[row][column]==card)

{

if(card%3=0)c='\n'

elsec='\t'

printf("%5sof%-8s%c",hua_se[row],dight[column],c)

}

}

}

}

}

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