说明: 50名队员围成一圈，顺序排号，从第一个开始报数(从1到7报数)，凡报到3和7的人退出圈子，报到7后，剩下的人再继续从1到7报数，编程实现最后留下的是原来第几号队员

#include <stdio.h>

size_t josephus( size_t n, size_t a, size_t b )
{
    size_t index = 0;
    for( size_t i=0; i!=n; ++i )
        index = (index+(i%2!=n%2?a:b))%(i+1);
    return index;
}

int main( void )
{
    printf( "%zu\n", josephus(50,3,4)+1 ); // 输出 40
}

/*
链表实现方式
*/
#include<stdio.h>
#include<stdlib.h>
typedef struct people
{
    int card;
    struct people *next;
};
void main()
{
    int i;
    int sum = 1;
    struct people *p,*q,*s,*m;//p表示循环输入的那个变量，q表示开始节点。
    p = q = (struct people *)malloc(sizeof(struct people));
    p->card = 1;
    p->next = (struct people *)malloc(sizeof(struct people));
    p = p->next;
    for(i = 0;i < 49;i++)
    {
        p->card = i+2;
        if(i != 48)
        {
            p->next = (struct people *)malloc(sizeof(struct people));
            p = p->next;
        }
        else
            p->next = q;
    }
    p = q;//p重新指向开始结点
    while(p != p->next)
    {
        if(sum == 3)
        {
            m = p;
            p = p->next;
            s->next = p;
                        sum++;
            free(m);
            m = NULL;
        }
                else if(sum==7)
                {
                        m = p;
            p = p->next;
            s->next = p;
                        sum = 1;
            free(m);
            m = NULL;
                }
        else
        {
            s = p;
            p = p->next;
            sum++;
        }
    }
    printf("50名队员围成一圈，顺序排号，从第一个开始报数(从1到7报数)，凡报到3和7的人退出圈子，报到7后，剩下的人再继续从1到7报数，最后留下的是原来第%d号队员\n",p->card);
    free(p);
    p = NULL;
}