UVA 133 (13.07.07)

The Dole Queue

In a serious attempt to downsize (reduce) the dole queue, The NewNational Green Labour Rhinoceros Party has decided on the followingstrategy. Every day all dole applicants will be placed in a largecircle, facing inwards. Someone is arbitrarily chosen as number 1,and the rest are numbered counter-clockwise up to N (who will bestanding on 1's left). Starting from 1 and moving counter-clockwise,one labour official counts off k applicants, while another officialstarts from N and moves clockwise, counting m applicants. The two whoare chosen are then sent off for retraining; if both officials pickthe same person she (he) is sent off to become a politician. Eachofficial then starts counting again at the next available person andthe process continues until no-one is left. Note that the two victims(sorry, trainees) leave the ring simultaneously, so it is possible forone official to count a person already selected by the other official.

Input

Write a program that will successively read in (in that order) thethree numbers (N, k and m; k, m > 0, 0 < N < 20) and determinethe order in which the applicants are sent off for retraining. Eachset of three numbers will be on a separate line and the end of datawill be signalled by three zeroes (0 0 0).

Output

For each triplet, output a single line of numbers specifying the orderin which people are chosen. Each number should be in a field of 3characters. For pairs of numbers list the person chosen by thecounter-clockwise official first. Separate successive pairs (orsingletons) by commas (but there should not be a trailing comma).

Sample input

10 4 3
0 0 0

Sample output

4 8, 9 5, 3 1, 2 6, 10, 7

where represents a space.

双向约瑟夫环

有个注意点是: 两个方向的遍历是同时进行了的, 不能先进行一个 然后就删除掉

要先同时打印, 然后同时删除~

代码如下:

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

typedef struct Pos{
    int num;
    struct Pos *last;
    struct Pos *next;
}pos;

//建表:
pos *creat(int n) {
    int i;
    pos *head, *p, *now;
    now = (pos*)malloc(sizeof(pos));
    head = now;
    p = now;
    head->num = 1;
    for(i = 2; i <= n; i++){
        now = (pos*)malloc(sizeof(pos));
        now->num = i;
        p->next = now;
        now->last = p;
        p = now;
    }
    p->next = head;
    head->last = p;
    return head;
}

int main() {
    //输入
    int i, N, k, m, flag;
    pos *head, *p, *pp, *pm, *pk;
    
    while(scanf("%d %d %d", &N, &k, &m) != EOF && N && k && m) {
        head = creat(N);
        pk = head;
        pm = head->last;
        while(head != NULL) {
            flag = 0;
            for(i = 1; i < k; i++)
                pk = pk->next;
            for(i = 1; i < m; i++)
                pm = pm->last;
            if(pk->num == pm->num) {
                printf("%3d",pk->num);
                p = pk;
                pk = pk->next;
                pm = pm->last;
                pk->last = pm;
                pm->next = pk;
                if(p->num == head->num) {
                    if(p->num == pk->num)
                        head = NULL;
                    else
                        head = head->next;
                }
                free(p);
                flag = 1;
            }
            else {
                printf("%3d%3d", pk->num, pm->num);
                p = pk;
                pp = pk->last;
                pk = pk->next;
                pp->next = pk;
                pk->last = pp;
                if(p->num == head->num)
                    head = head->next;
                free(p);
                p = pm;
                pp = pm->next;
                pm = pm->last;
                pp->last = pm;
                pm->next = pp;
                if(pk->num == p->num)
                    pk = pk->next;
                if(p->num == head->num) {
                    if(p->num == pm->num)
                        head = NULL;
                    else
                        head = head->next;
                }
                free(p);
                flag = 1;
            }
            if(flag && head != NULL)
                printf(",");
        }
        printf("\n");
    }
    return 0;
}