约瑟夫环（出列问题）

最早大学讲到这个问题时，就是学习C语言到数组的时候，只需要用数组和指针就可以解决，好像解决的方法还很多。
主要需要解决两个问题：
1.指针循环递增到数组结尾时，需要跳转到数组开头。
2.出列后，数组中相应的值设为0。
其实还要考虑到一个小问题，比如：总人数100人，数到13出列，如果刚开始指针在数组的第一位，数到13时，就对应数组的第14个数，这肯定不行。所以数组设为a[101]，a[0]不计入人数，从a[1]至a[100]计入1至100的数字。

#include 
#include 

#define NUM 100 //The total number of people who stand around in a circle .
#define REP 15  //The counting number the REP-th person who will be executed .

unsigned jos(unsigned n,unsigned rec)
{
    unsigned a[n+1],b[n],i,j=1;   //From a[1] to a[n] is the number of 1 to n; and the numbers from b[0] to b[n-1] is the result of Josephus permutation.
    unsigned *pio=a;
    a[0]=999;              //The count begins at 0(zero),and the next count is 1.a[0] can be assigned any value beyond 0 and NUM.
    for(i=1;i<=n;i++){    //Assign the value from 1 to n to a[1]......a[n] respectively .
	    pio++;
	    *pio=i;
    }
   
    pio=&a[0];         //Initialize *pio.

    for(i=0;iNUM){
	   printf("ERROR !\nThe counting number is bigger than the total number of people !\n ");
	   return 1;             
	}

	int mp,i=0;

	mp = jos(NUM,REP);
	printf("\nThe final number is %d.\n\n\n",mp);
	end = clock();   //Time off .
	
    printf("time=%.4f\n", (double)(end - start) / CLOCKS_PER_SEC);	
	return 0;
}