㈠ c語言 紙牌大戰
每個游戲者生成52個數的數組，然後每個游戲者生成兩個指針：棧頭、棧尾、當前正在出牌的指針，非數組變數包括一個出牌計數。
實現的方法：在一個循環里反復的出牌（移動當前正在出牌的指針），贏得則將出牌數組完全移到贏者的數組中。
問題：贏者以什麼順序收牌？先收自己的還是先收對方的，先收先發的還是先收後發的？如果先收自己，先收先發，那麼是收自己先發-收自己後發-收對方先發-收對方後發，還是收自己先發-收對方先發-收自己後發-收對方後發？
不過，概述有了，實現就很容易了
㈡ C語言程序設計課程設計撲克牌游戲，怎麼做
#include<conio.h>
#include<time.h>
#include<stdio.h>
#include<stdlib.h>
int jisuan(int);
int comptotal;
char s1[]="A234567890JQK";
//char s2[4][5]={"紅桃","黑桃","草花","方塊"};
char s2[4]={3,4,5,6};
int poke[52];
int ch;
int win=0;
int computer[5],user[5];
int usertotal;
int users;
int k;
int main()
{ void xipai(int poke[]);
void ai();
int i,j;
////////////////////////////////////////////////////////上面是變數和聲明
printf("\n這是簡單的廿一點游戲:\n");
for(i=0;i<52;i++)
{
if(i%13==0)putchar('\n');
poke[i]=i;
printf("%c%c%c ",s2[i/13],s1[i%13]=='0'?'1':' ',s1[i%13]);
}
putchar('\n');
/////////////////////////////////////////////////////////主代碼
k=0;
xipai(poke);
while(ch!=27)
{ comptotal=0;
usertotal=0;
if(k>=42)
{
printf("\n剩餘牌數不足十張,重新洗牌");
xipai(poke);
k=0;
}
printf("\n\n\n\n\n\n\n\n新局開始:\n");
printf("現在共有牌%2d張\n",52-k);
if(win==0)
{
computer[0]=k++;
user[0]=k++;
printf("\n電腦做莊,要牌:");
ai();
}
else
{
printf("\n玩家做莊,要牌:\n\t回車要牌\n\t空格過牌");
user[0]=k++;
computer[0]=k++;
}
printf("\n玩家開始要牌:\n");
usertotal=jisuan(poke[user[0]]);
printf("%c%c%c 共%2d點\t",s2[poke[user[0]]/13],s1[poke[user[0]]%13]=='0'?'1':' ',s1[poke[user[0]]%13],usertotal);
users=0;
ch=1;
while(ch!=32&&users<4)
{
ch=getch();
switch(ch)
{
case 27:
goto end;
break;
case 32:
break;
case 13:
user[++users]=k;
usertotal+=jisuan(poke[user[users]]);
printf("\b\b\b\b\b\b\b\b\b%c%c%c 共%2d點\t",s2[poke[k]/13],s1[poke[k]%13]=='0'?'1':' ',s1[poke[k]%13],usertotal);
k++;
if(usertotal>=21)ch=32;
break;
default:
break;
}
}
if(win==1)
{
printf("\n電腦開始要牌:\n");
ai();
}
printf("\n\n\n玩家的點數是%2d",usertotal);
printf("\n電腦的點數是%2d",comptotal);
printf("\n\n本局結算:");
if(comptotal>21&&usertotal<=21)
{
printf("\n\n電腦爆牌了");
win=1;
printf("\n恭喜,你贏了");
}
if(usertotal>21&&comptotal<=21)
{
printf("\n\n你爆牌了");
printf("\n下次小心點");
win=0;
}
if(usertotal>21&&comptotal>21)
{
printf("\n\n你們兩個，怎麼都這么不小心啊,都撐死了還要嗎");
}
if(usertotal<=21&&comptotal<=21)
{
if(usertotal>comptotal)
{
win=1;
printf("\n\n不錯，你贏了");
}
else if(usertotal<comptotal)
{
win=0;
printf("\n\n撐死膽大的，餓死膽小的，沒膽子，輸了吧");
}
else
printf("\n\n平了，算你走運");
}
getch();
}
end:
return 0;
}
void xipai(int poke[])
{
int y,tmp,i,j;
for(j=0;j<7;j++)
for(i=0;i<52;i++)
{
srand(time(0));
y=rand()%10;
tmp=poke[i];
poke[i]=poke[(y*i*i)%52];
poke[(y*i*i)%52]=tmp;
}
}
///////////////////////////////////////////////子函數
void ai()
{
int i;
comptotal=jisuan(poke[computer[0]]);
printf("\n%c%c%c 共%2d點\t",s2[poke[computer[0]]/13],s1[poke[computer[0]]%13]=='0'?'1':' ',s1[poke[computer[0]]%13],comptotal);
for(i=0;i<4;i++)
{
if(comptotal<17)
{
computer[i+1]=k++;
comptotal+=jisuan(poke[computer[i+1]]);
printf("\b\b\b\b\b\b\b\b\b%c%c%c 共%2d點\t",s2[poke[computer[i+1]]/13],s1[poke[computer[i+1]]%13]=='0'?'1':' ',s1[poke[computer[i+1]]%13],comptotal);
}
}
}
int jisuan(int i)
{int dian;
switch(i%13)
{
case 0:
case 10:
case 11:
case 12:
dian=1;
break;
default:
dian=i%13+1;
}
return dian;
}
㈢ C語言編程，紙牌游戲
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <dos.h>
#include <graphics.h>
#include <conio.h>
#define ESC 0x1b
struct card
{
char color;
int number;
int signin;
}a[52]={{3,2,1},{3,3,1},{3,4,1},{3,5,1},{3,6,1},{3,7,1},{3,8,1},{3,9,1},{3,10,1},{3,74,1},{3,81,1},{3,75,1},{3,65,1},
{4,2,1},{4,3,1},{4,4,1},{4,5,1},{4,6,1},{4,7,1},{4,8,1},{4,9,1},{4,10,1},{4,74,1},{4,81,1},{4,75,1},{4,65,1},
{5,2,1},{5,3,1},{5,4,1},{5,5,1},{5,6,1},{5,7,1},{5,8,1},{5,9,1},{5,10,1},{5,74,1},{5,81,1},{5,75,1},{5,65,1},
{6,2,1},{6,3,1},{6,4,1},{6,5,1},{6,6,1},{6,7,1},{6,8,1},{6,9,1},{6,10,1},{6,74,1},{6,81,1},{6,75,1},{6,65,1}},b[52];
char s[10];
int d[52];
fan(int n)
{
if(a[n].signin==0)
a[n].signin=1;
else a[n].signin=0;
return;
}
suiji()
{
int t=0,j,i=0,struction;
for(i=0;i<=51;i++)
d[i]=-1;
i=0;
randomize();
while(i<52)
{
struction=random(52);
for(j=0;j<i;j++)
{
if(d[j]==struction)
{
t=1;
break;
}
}
if(t==0)
{
d[i]=struction;
i++;
}
else t=0;
}
return;
}
card(int n)
{
int y,x=n%13,x1=d[n]%13,y1;
char s1[2],r[2];
y=(n-x)/13;
setcolor(15);
line(49*x,80*y+42,49*x,80*y+117);
line(49*x+47,80*y+42,49*x+47,80*y+117);
line(49*x,80*y+42,49*x+47,80*y+42);
line(49*x,80*y+117,49*x+47,80*y+117);
for(y1=1;y1<75;y1++)
{
setcolor(15);
line(49*x+1,80*y+y1+42,49*x+45,80*y+y1+42);
}
if(a[n].signin==1)
{
setcolor(1);
if(a[d[n]].color==3||a[d[n]].color==4)
setcolor(RED);
sprintf(s1,"%c",a[d[n]].color);
outtextxy(49*x+3,80*y+45,s1);
if(x1<9)
sprintf(r,"%d",a[d[n]].number);
else sprintf(r,"%c",a[d[n]].number);
outtextxy(49*x+3,80*y+54,r);
}
else against(n);
setcolor(WHITE);
return;
}
against(int n)
{
int y,y1,x=n%13;
y=(n-x)/13;
for(y1=1;y1<75;y1++)
{
setcolor(BLUE);
line(49*x+1,80*y+y1+42,49*x+45,80*y+y1+42);
}
setcolor(15);
return;
}
draw(int n,int m)
{
int i4;
setcolor(YELLOW);
sprintf(s,"Base: %d",n);
outtextxy(150,420,s);
if(d[m-1]%13<9&&d[m-1]%13>=0)
sprintf(s,"Card: %c %d",a[d[m-1]].color,a[d[m-1]].number);
else sprintf(s,"Card: %c %c",a[d[m-1]].color,a[d[m-1]].number);
outtextxy(150,430,s);
setcolor(15);
for(i4=0;i4<52;i4++)
{
card(i4);
}
return;
}
frame(int n)
{
int y,x=n%13;
y=(n-x)/13;
setcolor(RED);
line(49*x,80*y+42,49*x,80*y+117);
line(49*x+47,80*y+42,49*x+47,80*y+117);
line(49*x,80*y+42,49*x+47,80*y+42);
line(49*x,80*y+117,49*x+47,80*y+117);
setcolor(15);
return;
}
huatu(int i,int j)
{
int n,m,i1=i,b,b2,j1=j,sign=0,tar=0,i2=2,k=1,p,sign1=0;
char u='',u1='',c[7][8],chh;
setbkcolor(3);
while(1)
{
loop:
sprintf(c[0],"File");
sprintf(c[1],"Option");
sprintf(c[2],"Help");
sprintf(c[3],"New");
sprintf(c[4],"Exit");
sprintf(c[5],"Auto");
sprintf(c[6],"Manual");
while(i2<=52)
{
k=1;
while(i2*k<=52)
{
setcolor(9);
for(n=0;n<=18;n++)
line(0,n,639,n);
setcolor(15);
outtextxy(290,10,"Card Game");
setcolor(YELLOW);
for(n=19;n<=40;n++)
line(0,n,639,n);
for(m=0;m<=2;m++)
{
setcolor(0);
line(58*m+3,22,58*m+3,38);
line(58*m+3,38,58*m+58,38);
line(58*m+58,22,58*m+58,38);
line(58*m+3,22,58*m+58,22);
outtextxy(8+58*m,27,c[m]);
if(m==i1)
{
setcolor(0);
for(n=0;n<=16;n++)
line(58*m+3,22+n,58*m+58,22+n);
setcolor(YELLOW);
outtextxy(8+58*m,27,c[m]);
}
if(kbhit())
{
chh=getch();
if(chh==0x1b)
{
u='';
u1='';
goto loop;
}
}
}
if(u!='N')
{
sign1=1;
break;
}
if(u=='N')
{
if(u1!='A'&&u1!='M')
{
p=i2*k;
draw(i2,p);
sign1=1;
break;
}
if(u1=='A'||u1=='M')
{
p=i2*k;
draw(i2,p);
sign1=0;
fan(i2*k-1);
frame(i2*k-1);
if(u1=='M')
{
outtextxy(150,440,"Press any key to continue!");
getch();
}
cleardevice();
k++;
}
}
}
if(sign1==1)break;
i2++;
}
if(i2>=52)
{
setcolor(9);
for(n=0;n<=18;n++)
line(0,n,639,n);
setcolor(15);
outtextxy(290,10,"Card Game");
setcolor(YELLOW);
for(n=19;n<=40;n++)
line(0,n,639,n);
outtextxy(150,420,"Press any key to continue!");
for(m=0;m<=2;m++)
{
setcolor(0);
line(58*m+3,22,58*m+3,38);
line(58*m+3,38,58*m+58,38);
line(58*m+58,22,58*m+58,38);
line(58*m+3,22,58*m+58,22);
outtextxy(8+58*m,27,c[m]);
if(m==i1)
{
for(n=0;n<=16;n++)
line(58*m+3,22+n,58*m+58,22+n);
setcolor(YELLOW);
outtextxy(8+58*m,27,c[m]);
}
} /*列印一級菜單*/
getch();
cleardevice();
setcolor(9);
for(n=0;n<=18;n++)
line(0,n,639,n);
setcolor(15);
outtextxy(290,10,"Card Game");
setcolor(YELLOW);
for(n=19;n<=40;n++)
line(0,n,639,n);
for(m=0;m<=2;m++)
{
setcolor(0);
line(58*m+3,22,58*m+3,38);
line(58*m+3,38,58*m+58,38);
line(58*m+58,22,58*m+58,38);
line(58*m+3,22,58*m+58,22);
if(m==i1)
{
for(n=0;n<=16;n++)
line(58*m+3,22+n,58*m+58,22+n);
setcolor(YELLOW);
}
outtextxy(8+58*m,27,c[m]);
}
draw(52,52);
}
if(u1=='M'||u1=='A')
{
u1='';
u='';
}
b=getch();
while(1)
{
if(b==100)
{
i1=(i1+1)%3;
break;
}
if(b==97)
{
if(i1==0)
{
i1=2;
break;
}
else
{
i1=i1-1;
break;
}
}
/* if(b==0x1b)
{
closegraph();
exit(1);
} */
if(b==13)
{
while(1)
{
if(i1!=2)
{
if(sign!=1)
{
setcolor(YELLOW);
for(m=1;m<=2;m++)
{
line(58*i1+3,38+15*(m-1),58*i1+3,53+15*(m-1));
line(58*i1+3,53+15*(m-1),58*i1+58,53+15*(m-1));
line(58*i1+58,38+15*(m-1),58*i1+58,53+15*(m-1));
line(58*i1+3,38+15*(m-1),58*i1+58,38+15*(m-1));
setcolor(3);
for(n=0;n<=13;n++)
{
line(58*i1+4,39+15*(m-1)+n,58*i1+57,39+15*(m-1)+n);
}
if(m==j1+1)
{
setcolor(3);
for(n=0;n<=13;n++)
{
line(58*i1+4,39+15*(tar)+n,58*i1+57,39+15*(tar)+n);
}
setcolor(BLUE);
for(n=0;n<=13;n++)
{
line(58*i1+4,39+15*(j1)+n,58*i1+57,39+15*(j1)+n);
}
tar=j1;
}
setcolor(YELLOW);
}
setcolor(YELLOW);
if(i1==0)
{
outtextxy(8+58*i1,43,c[i1+3]);
outtextxy(8+58*i1,58,c[i1+4]);
}
else
{
outtextxy(8+58*i1,43,c[i1+4]);
outtextxy(8+58*i1,58,c[i1+5]);
}
}
if(sign==1)break;
setcolor(YELLOW);
while(1)
{
b2=getch();
if(b2==115)
{
j1=(j1+1)%2;
break;
}
if(b2==119)
{
if(j1==1)
{
j1=0; break;
}
if(j1==0)
{
j1=1; break;
}
}
if(b2==0x1b)
{
sign=1;
j1=0;
break;
}
if(b2==13)
{
if(i1==0&&j1==1)
{
closegraph();
exit(1);
}
if(i1==0&&j1==0)
{
u='N';
suiji();
for(n=0;n<52;n++)
{
a[n].signin=1;
}
n=0;
}
if(i1==1&&j1==0)
{
u1='A';
if(u!='N') u1='';
}
if(i1==1&&j1==1)
{
u1='M';
if(u!='N') u1='';
}
sign=1;
j1=0;
break;
}
else continue;
}
}
else
{
sign=1;
outtextxy(100,100,"Copyright");
circle(180,104,5);
outtextxy(177,101,"C");
getch();
break;
}
}
}
else break;
if(sign==1)
{
sign=0;
break;
}
}
cleardevice();
i2=2;
k=1;
}
}
main()
{
int k,gm=2,gd=9;
initgraph(&gd,&gm,"");
huatu(0,0);
getch();
}
㈣ 用C語言編程撲克牌搓點游戲，急！
給你修改好了。20分，有點少了，呵呵。
/*游戲：撲克牌搓點游戲
規則：您將隨機抽取其中兩張和電腦進行對抗，
2張牌相加，個位大的勝出，其中對子比單牌大，
若都是對子，對子大的勝出。
您可以根據提示下注，起始資金均為1000元，
當一方財產小於0時，宣布破產，另一方勝出。
作者：於吉祥
日期：20090220
*/
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define MAX_M 1000
enum colour
{
HEI = 0,
HONG,
MEI,
FANG,
};
void display(int number , int sign)
{
printf("%s\n","╭——╮");
switch(number)
{
case 0:
printf("%s\n","│0 │");
break;
case 1:
printf("%s\n","│A │");
break;
case 2:
printf("%s\n","│2 │");
break;
case 3:
printf("%s\n","│3 │");
break;
case 4:
printf("%s\n","│4 │");
break;
case 5:
printf("%s\n","│5 │");
break;
case 6:
printf("%s\n","│6 │");
break;
case 7:
printf("%s\n","│7 │");
break;
case 8:
printf("%s\n","│8 │");
break;
case 9:
printf("%s\n","│9 │");
break;
case 10:
printf("%s\n","│10 │");
break;
case 11:
printf("%s\n","│J │");
break;
case 12:
printf("%s\n","│Q │");
break;
case 13:
printf("%s\n","│K │");
break;
default:
printf("error");
break;
}
// printf("%s\n","│ │");
printf("%s","│ ");
switch(sign)
{
case HEI:
printf("%c",06);
break;
case HONG:
printf("%c",03);
break;
case MEI:
printf("%c",05);
break;
case FANG:
printf("%c",04);
break;
default:
break;
}
printf("%s\n"," │");
printf("%s\n","│ │");
printf("%s\n","╰——╯");
}
int compare(int x[] , int y[])
{
int sign,a,b;
if((x[0] == x[1])&&(y[0] == y[1]))
{
if(x[0] == y[0])
sign = 0;
else if(x[0] > y[0])
sign = 1;
else
sign = -1;
}
else if(x[0] == x[1])
sign = 1;
else if(y[0] == y[1])
sign = -1;
else
{
a = (x[0]+x[1])%10;
b = (y[0]+y[1])%10;
if(a == b)
sign = 0;
else if(a > b)
sign = 1;
else
sign = -1;
}
return sign;
}
void main()
{
int i,chip;
int Per[2],Com[2];
int Mon_Per = MAX_M , Mon_Com = MAX_M;
int Colour_Per[2] , Colour_Com[2];
printf("撲克牌搓點游戲\n");
printf("---------------\n");
system("pause");
while(1)
{
system("cls");
if(Mon_Per <=0)
{
printf("您已身無分文，游戲退出！");
break;
}
if(Mon_Com <=0)
{
printf("電腦已經被你贏光了，恭喜你獲勝！");
break;
}
srand( time(NULL) );
for(i=0;i<2;i++)
{
Per[i] = rand()%13+1;
Colour_Per[i] = rand()%4;
Com[i] = rand()%13+1;
Colour_Com[i] = rand()%4;
}
printf("當前余額：你（%d）,電腦（%d）\n",Mon_Per,Mon_Com);
// printf("牌已經抽取，你抽到的牌為：%d %d\n",Per[0],Per[1]);
printf("牌已經抽取，你抽到的牌為：\n");
for(i=0;i<2;i++)
display(Per[i],Colour_Per[i]);
printf("請下註：");
RET: scanf("%d",&chip);
if(chip>500)
{
printf("最大可下注為500，請重新下注！\n");
goto RET;
}
Mon_Per -= chip;
Mon_Com -= chip;
if(Mon_Per < 0)
{
printf("您的余額不足，請重新下注！");
Mon_Per += chip;
Mon_Com += chip;
goto RET;
}
if(Mon_Com < 0)
{
printf("電腦余額不足，請重新下注！");
Mon_Per += chip;
Mon_Com += chip;
goto RET;
}
switch(compare(Per,Com))
{
case 0:
printf("平局！");
Mon_Per += chip;
Mon_Com += chip;
break;
case 1:
printf("你贏了！");
Mon_Per += 2*chip;
break;
case -1:
printf("電腦贏了！");
Mon_Com += 2*chip;
break;
default:
printf("系統出錯！");
break;
}
// printf("電腦抽到的牌為：%d %d\n",Com[0],Com[1]);
printf("電腦抽到的牌為：\n");
for(i=0;i<2;i++)
display(Com[i],Colour_Com[i]);
system("pause");
}
printf("游戲結束！");
system("pause");
}
㈤ c語言程序設計撲克牌游戲
定義一個結構類型表示一張牌，結構包含3個成員，第一個成員char：取值2，3~K，A表示牌名字，第二個成員int:取值2~14表示牌真實大小。第三個成員:結構鏈表指針。
寫一個初始化函數，定義52大小的結構數組，成員值初值分別和牌對應，遍歷數組並將每個元素的鏈表指針依次指向下一個元素地址。這樣得到一個初始鏈表。（相當於一盒新牌）
所有涉及隨機數都用rand函數，洗牌分四份就是循環取隨機數m=1~n，n是隨循環自減，初值52，直到n變成0。每隨一次循環就從初始鏈表中遍歷取出對應第m個節點，並從初始鏈表中將這個節點斷開（既前一個節點指針直接指向後一個節點指針）。每取13張就組成一個新的鏈表。這樣獲得4個新鏈表分別表示4個玩家。
最後出牌就是分別遍歷自己的鏈表，利用循環取牌比較結構數值大小。（取出的牌要從鏈表斷開和上面一樣，你把取出節點寫成獨立函數就能反復使用）。
㈥ 求c語言的「紙牌游戲」代碼
貪心演算法，首先計算平均值，然後從左往右掃描，不夠平均值的從右邊拿，次數+1；超過平均值的往右邊放，次數+1。#include
int
main(){
int
n,i,move=0;
long
a[101],num=0;
scanf("%d",&n);
for
(i=0;i
num){
a[i+1]+=(a[i]-num);
move++;
}
if
(a[i]
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㈦ C語言設計憋七紙牌游戲
/* 紙牌模擬程序 */
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
struct card
{ char suit[2];
char face[3];
};
typedef struct card CARD;
void initcard(CARD deck[],char s[][2],char f[][3]);
void shuffle(CARD deck[]);
void print(CARD deck[]);
int main()
{ CARD deck[52];
char s[4][2]={"\003","\004","\005","\006"};
char f[13][3]={"A","2","3","4","5","6","7","8","9","10","J","Q","K"};
initcard(deck,s,f);
srand(time(NULL));
shuffle(deck);
print(deck);
system("pause");
return 0;
}
void initcard(CARD deck[],char s[][2],char f[][3])
{ int i;
for(i=0;i<52;i++)
{ strcpy(deck[i].suit,s[i/13]);
strcpy(deck[i].face,f[i%13]);
}
}
void shuffle(CARD deck[])
{ int i,j;
CARD temp;
for(i=0;i<52;i++)
{ j=rand()%52;
if(j!=i)
{ temp=deck[i];
deck[i]=deck[j];
deck[j]=temp;
}
}
}
void print(CARD deck[])
{ int i;
for(i=0;i<52;i++)
{ printf("%2s--%2s",deck[i].suit,deck[i].face);
printf("%c",(i+1)%4? '\t' : '\n');
}
}
㈧ C語言 紙牌游戲——比大小 題目描述 蔥蔥跟巴豆玩游戲。游戲規則很簡單，就是比大小。
#include <iostream>
#include <memory.h>
#include <algorithm>
using namespace std;
#define N 11
int main()
{
int cc[N] ;
bool b[210] ;
int i, j ;
int x ;
int count ;
bool flag ;
while(true)
{
memset(b, false, sizeof(b)) ;
count = 0 ;
for(i = 0; i < N; i++)
cin >> cc[i] ; //輸入蔥蔥的牌
for(i = 0; i < N; i++) //輸入巴豆的牌
{
cin >> x ;
b[x] = true ;
}
sort(cc, cc+N) ; //將蔥蔥手裡的牌排序
j = 1 ;
for(i = 0; i < N; i++)
{
x = cc[i] ;
flag = false ;
for( ; j < x && !flag; j++)
{
if(b[j])
{
flag = true ;
b[j] = false ; //j已經被比較過
count ++ ;
}
}
}
cout << count << endl ;
}
return 0;
}
//結果不一定完全正確。
㈨ 求c語言 翻紙牌游戲
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<conio.h>
intmain()
{
inta,b;
inti,p;
intoc=0;
intcard[6]={0,0,0,0,0,0};
intopened[6]={0,0,0,0,0,0};
srand(time(0));
for(i=1;i<=6;i++)
{
p=rand()%6;
while(card[p])p=rand()%6;
card[p]=(i+1)/2;
}
printf("按任意鍵開始");
getch();
while(oc<3)
{
do
{
system("cls");
for(inti=0;i<6;i++)
if(opened[i])
printf("%d",card[i]);
else
printf("*");
printf(" ");
printf("輸入你第一張牌的位置:");
scanf("%d",&a);
}while(!(0<a&&a<=6&&!opened[a-1]));
do
{
system("cls");
for(inti=0;i<6;i++)
if(opened[i]||i==a-1)
printf("%d",card[i]);
else
printf("*");
printf(" ");
printf("輸入你第二張牌的位置:");
scanf("%d",&b);
}while(!(0<b&&b<=6&&!opened[b-1]&&b!=a));
system("cls");
for(inti=0;i<6;i++)
if(opened[i]||i==a-1||i==b-1)
printf("%d",card[i]);
else
printf("*");
printf(" ");
printf("按鍵繼續");
getch();
if(card[a-1]==card[b-1])
{
oc++;
opened[a-1]=1;
opened[b-1]=1;
}
}
system("cls");
printf("恭喜你成功了");
return0;
}
運行效果：
㈩ C語言實現紙牌游戲（將不含有大小王的紙牌進行隨機發派給兩個人並對紙牌進行比較）
居然都沒分數獎勵的啊。。
用一個數組 A[52] 紅桃 黑桃 梅花 方片 倍數分別為0,1,2,3 則黑桃 1 為 1*13 +0 =A[13] 方片2數值為 3*13 + 1 = A[40] 這樣從A[0] -A[51]表示這52張牌。
然後A[10] = 0,1,2 初始化全為 0 表示還沒發，每次 用隨機數除以當前總數 取余 來隨機拍一張牌，例如隨機0-52的數字 比如 48表示A[47]發出去，發給甲 A[47]=1 乙為2，以後每次隨機一個0-52的數字 用加法，當當前牌的屬性為0時加1否則不加，加到隨機數為止，這樣可以相當於一直在沒有發出去的牌做隨機，加的和超過52取余
做52次發牌操作後結束，每次從值為1 和2的中隨機一個序列號用序列號除以13取余比大小。
思路全說了，代碼天天上班寫就懶得給你寫了