uva133

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

这个题就是模拟，其实挺容易的，不过我由于手残，做了一个小时···而且以前都不知道memset()初始化是按字节来的，这次算是学习了，这里得注意的就是数组的下标得弄明白，宁愿写得细致一点也不要为了图快就直接写，这样不好调试噶···

#include<cstdio>#include<cstdlib>#include<cstring>#define LOCALint p[25];int main(){    #ifdef LOCAL        freopen("input.txt","r",stdin);        freopen("output.txt","w",stdout);    #endif     int n,k,m,p1,p2;    while(scanf("%d %d %d",&n,&k,&m)&&n&&m&&k){        memset(p,0,sizeof(p));        int num=n,i=1,j=n;        while(num){             for(int x=0;x<k;i++){                if(i>n)i%=n;                if(p[i]==0)x++;             }              i--;             for(int y=0;y<m;j--){                 if(j==0)j=n;                 if(p[j]==0)y++;                         }             j++;             if(i==j){  printf("%3d",j); p[j]=1;num--; }             else{printf("%3d%3d",i,j);p[i]=1;p[j]=1;num-=2;}             if(num!=0)printf(",");             else printf("\n");        }    }    return 0;}