Coder Buct 1072: 数据结构作业（可选）-- 约瑟环问题

/*NO.1 (O(n) )算法*/#include <stdio.h>
#include <math.h>int main()
{int m, r;int i, j, s = 0;scanf("%d %d",&m, &r);for(i = 1; i <= m; i++)s = (s + r) % i;printf("%d\n",s + 1);return 0 ;
}/*NO.2 数组模拟*/#include <stdio.h>
#include <stdlib.h>
#define MAX 10001/*利用：求余特性,(i+1)%m...*/
int main()
{int a[MAX] = {0} ;int m, r, i, j, k, h;scanf("%d%d",&m, &r) ;k = 0;for(i = 1; i <= m; i++){for(j = 0 ; ; ){if(a[k] == 0){j++; if(j < r)   k = (k + 1) % m;else{a[k] = 1;printf("%d\n",k+1); /*输出正解*/ if(i != m)k = (k + 1) % m;break;}}elsek = (k+1) % m ;}}printf("%d\n",k+1) ;  return 0 ;
}/*NO.3循环链表模拟*/#include <stdio.h>
#include <stdlib.h>typedef struct list
{int num;struct list *next;
}LIST;int main()
{LIST *head = NULL, *p, *q;long i, j, k, m, r;scanf("%ld%ld", &m, &r) ;if(m <= 1){printf("1");return 0;}/*建立循环链表*/p = (LIST *)malloc (sizeof(LIST));p->num = 1;p->next = NULL;head = p;q = p;for(i = 2; i <= m; i++){p = (LIST *)malloc (sizeof(LIST));p->num = i;p->next = NULL;q->next = p;q = p;}q->next = head ; /*将最后一个结点的指向头，这样就构成了循环链表*/p = head ;while (q->next != q){for(i = 1; i < r; i++){q = p;p = p->next;}q->next = p->next ;p = q->next ;}printf("%d\n", p->num);           return 0 ;
}