编写"怪异的洗牌"时洗牌只能洗一次了

#include <stdio.h>

// 题目中提到的shift和flip，都需要用到反转算法，故而将其独立出来
void reverse( unsigned cards[], size_t length )
{
    for( size_t i=0,j=length; i+1<j; ++i )
    {
        unsigned t = cards[i];
        cards[i] = cards[--j];
        cards[j] = t;
    }
}

// 从某个位置分成两半，然后相交换
void shift( unsigned cards[], size_t length, size_t pos )
{
    reverse( cards, pos );
    reverse( cards+pos, length-pos );
    reverse( cards, length );
}

// 把后一半（如果总数是奇数，就是(n-1)/2）牌翻转过来
void flip( unsigned cards[], size_t length )
{
    reverse( cards+(length+1)/2, length/2 );
}

// 先进行一次shift再进行一次flip为一次完整的洗牌
void shuffle( unsigned cards[], size_t length, size_t pos )
{
    shift( cards, length, pos );
    flip( cards, length );
}

// 输出所有牌面
void show( unsigned index, const unsigned cards[], size_t length )
{
    printf( "第%u次翻牌后的结果为: ", index );
    for( size_t i=0; i!=length; ++i )
        printf( "%u%c", cards[i], " \n"[i+1==length] );
}

int main( void )
{
    enum {
        n = 9,   // 牌的数目
        m = 7,    // 切牌位置
        count = 5 // 洗牌次数
    };
   
    unsigned cards[n];
    for( unsigned i=0; i!=n; ++i )
        cards[i] = i+1;
   
    for( unsigned i=0; i!=count; ++i )
    {
        show( i, cards, n );
        shuffle( cards, n, m );
    }   
}