UVA-133 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 counterclockwise 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 counterclockwise official first. Separate successive pairs (or singletons) 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

分析：两边的约瑟夫环问题。

Source：

#include<stdio.h>int vis[20],n;int move(int pos,int flag,int step){    int i;    for(i=1;i<=step;i++)    {        do{            pos=(pos+flag+n-1)%n+1;        /*防止出现0和负数*/        }while(vis[pos-1]);    }    return pos;}int main(){    int left,p1,p2;    int k,m,i;    while(scanf("%d%d%d",&n,&k,&m)!=EOF)    {        if(n==0&&k==0&&m==0)            return 0;        for(i=0;i<n;i++)            vis[i]=0;        p1=n;        p2=1;        left=n;        while(left)        {            p1=move(p1,1,k);            p2=move(p2,-1,m);            left--;            printf("%3d",p1);            if(p1!=p2)            {                printf("%3d",p2);                left--;            }            if(left)                printf(",");            else                printf("\n");            vis[p1-1]=vis[p2-1]=1;        }    }    return 0;}