C语言实现洗牌与发牌游戏
本文实例为大家分享了C语言实现洗牌与发牌游戏的具体代码,供大家参考,具体内容如下
普通版
#include<stdio.h> #include<stdlib.h> #include<time.h> #define SUITS 4 #define FACES 13 #define CARDS 52 void shuffle(int wDeck[][FACES]); void send(int wDeck[][FACES], const char* wSuit[], const char* wFace[]); void main() { int deck[SUITS][FACES]={0}; const char* suit[SUITS]={"Heart","Diamond","Club","Spade"}; const char* face[FACES]={"Ace","Two","Three","Four","Five","Six","Seven","Eight","Nine","Ten","Jack","Queen","King"}; srand(time(NULL)); shuffle(deck); send(deck, suit, face); } void shuffle(int wDeck[][FACES]) { int row,column,card; for(card=1;card<=CARDS;card++) { row = rand() % SUITS; column = rand() % FACES; while(wDeck[row][column] != 0) { row = rand() % SUITS; column = rand() % FACES; } wDeck[row][column] = card; } } void send(int wDeck[][FACES], const char* wSuit[], const char* wFace[]) { int card,row,column; for(card=1;card<=CARDS;card++) { for(row=0;row<SUITS;row++) { for(column=0;column<FACES;column++) { if(wDeck[row][column] == card) { printf("%5s of %-8s%c",wFace[column],wSuit[row], card % 2 ==0 ? '\n' : '\t'); } } } } }
高性能版
//先按顺序填满52个格子,再产生随机格子相互交换 #include<stdio.h> #include<stdlib.h> #include<time.h> #define CARDS 52 #define FACES 13 struct card { const char* face; const char* suit; }; typedef struct card Card; void shuffle(Card *const wDeck); void send(const Card *const wDeck);//指针常量与数据常量均不可以改变,都用const限定,*前是限定指针,*后是限定数据 void main() { const char* wSuit[]={"Heart","Diamond","Club","Spade"};//指针数组是一个数组 const char* wFace[]={"Ace","Two","Three","Four","Five","Six","Seven","Eight","Nine","Ten","Jack","Queen","King"}; int i; Card deck[CARDS]; for(i=0;i<CARDS;i++) { deck[i].suit = wSuit[i / FACES]; deck[i].face = wFace[i % FACES]; } srand(time(NULL)); shuffle(deck); send(deck); } void shuffle(Card *const wDeck) { int i,j; Card temp; for(i=0;i<CARDS;i++) { j = rand() % CARDS; temp = wDeck[i]; wDeck[i] = wDeck[j]; wDeck[j] = temp; } } void send(const Card *const wDeck) { int i; for(i=0;i<CARDS;i++) { printf("%5s of %-8s%c",wDeck[i].face,wDeck[i].suit, (i+1) % 4 == 0 ? '\n' : '\t'); } }
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