UVA 133(循环链表)

C - The Dole Queue

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status

Description

 The Dole Queue 

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

Input

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

Output

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

Sample input

10 4 30 0 0

Sample output

 4  8,  9  5,  3  1,  2  6,  10,  7

where  represents a space.

题意：一个循环的圈，由两个人从头尾开始查找，分别将第m,第n的人删除，（两人是同时进行的），即是数完了才删除的，遇到相同的只输出一个，否则都输出，然后重新开始数，直至全部输完为止。直接用循环链表模拟即可，需要注意要删除的两个人相邻的情况，需要特殊处理，代码如下

#include<cstdio>#include<cstring>#include<algorithm>#define front Frontint N,M,K,num[21],next[21],front[21],ans[21];int main(){    while(~scanf("%d%d%d",&N,&M,&K)&&(M+N+K))    {        for(int i=0;i<N;i++)        {            num[i]=i+1;            next[i]=i+1;            front[i]=i-1;        }        front[0]=N-1;        next[N-1]=0;        int count=0,head=0,rear=N-1,t1=1,t2=1,p=N-1,q=0;///p,q,分别为head和rear的前一节点        while(count<N)        {            while(t1<M){p=head;head=next[head];t1++;}            while(t2<K){q=rear;rear=front[rear];t2++;}            t1=t2=1;            if(num[head]!=num[rear])            {                count+=2;                printf("%3d%3d",num[head],num[rear]);                if(count<N)putchar(',');            }            else///相等只打印一个            {                count+=1;                printf("%3d",num[head]);                if(count<N)putchar(',');            }            if(next[head]==rear)///head的后驱等于rear            {                next[p]=next[rear];                front[q]=front[head];                head=next[rear];                rear=front[head];            }            else if(front[head]==rear)///head的前驱等于rear(一定要注意判断这些情况）            {                next[front[rear]]=next[head];                front[next[head]]=front[rear];                head=next[head];                rear=front[rear];            }            else///head ,rear不相邻            {                next[p]=next[head];                front[next[head]]=front[head];                front[q]=front[rear];                next[front[rear]]=next[rear];                head=next[head];                rear=front[rear];            }        }        printf("\n");    }    return 0;}