用c语言写桥牌共有52张，不用大小王，按E,S,W,N的顺序把52张牌随机发给4个玩家，请列出每个玩家的发牌情况。

在C语言中，我们可以使用`<stdio.h>`、`<stdlib.h>`以及`<time.h>`库来实现这个功能。这里我们先定义牌的枚举和结构体，然后创建一个函数来打乱牌并分配给玩家。下面是一个简单的示例：

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

typedef enum { SPADE = 0, HEART, DIAMOND, CLUB } Suit;
enum Rank { ACE, TWO, THREE, ..., KING };

typedef struct {
    int rank;
    Suit suit;
} Card;

void shuffle(Card deck[], int count) {
    srand(time(0));
    for (int i = count - 1; i > 0; i--) {
        int j = rand() % (i + 1);
        Card temp = deck[i];
        deck[i] = deck[j];
        deck[j] = temp;
    }
}

void deal_cards(Card deck[], int players[4], int num_cards_per_player) {
    for (int i = 0; i < num_cards_per_player * 4; i++) {
        players[i / num_cards_per_player] = deck[i];
    }
}

int main() {
    const int NUM_CARDS = 52;
    Card deck[NUM_CARDS];

    // 初始化牌
    for (int i = 0; i < SPADE; i++) {
        for (Suit s = SPADE; s <= CLUB; s++) {
            deck[i++] = (Card){i + 1, s};
        }
    }

    // 洗牌
    shuffle(deck, NUM_CARDS);

    // 分配给四个玩家，每名玩家13张牌
    int player1[NUM_CARDS / 4], player2[NUM_CARDS / 4], player3[NUM_CARDS / 4], player4[NUM_CARDS / 4];
    deal_cards(deck, player1, NUM_CARDS / 4);
    deal_cards(deck, player2, NUM_CARDS / 4, NUM_CARDS / 4);
    deal_cards(deck, player3, NUM_CARDS / 4, NUM_CARDS / 4 * 2);
    deal_cards(deck, player4, NUM_CARDS / 4, NUM_CARDS / 4 * 3);

    // 输出每个玩家的牌
    printf("Player 1: ");
    for (int i = 0; i < NUM_CARDS / 4; i++) {
        printf("(%d of %s), ", player1[i].rank, to_suit(player1[i].suit));
    }
    printf("\n");

    // ...类似地打印其余三个玩家的牌

    return 0;
}

// 辅助函数将Suit转换为字符串
char* to_suit(Suit suit) {
    switch (suit) {
        case SPADE: return "Spades";
        case HEART: return "Hearts";
        case DIAMOND: return "Diamonds";
        case CLUB: return "Clubs";
    }
    return "";
}

注意这只是一个简化版本的示例，并未处理打印玩家牌的具体细节。你可以根据需要添加更多的错误检查和格式化输出。