L - The Dole Queue-蒲公英云

L - The Dole Queue

青旅半醒 2022-10-01 04:51 65阅读 0赞

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 pickthe 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). Note: The symbol ⊔ in the Sample Output below represents a space.

Sample Input
10 4 3
0 0 0

Sample Output
␣␣4␣␣8,␣␣9␣␣5,␣␣3␣␣1,␣␣2␣␣6,␣10,␣␣7

#include<stdio.h>
#include<string.h>
int main()
{ 
    int a[22], i, j, n, t, k, m, left, right;
    while(scanf("%d %d %d", &n, &k ,&m) == 3 && n && m && k)
    { 
        memset(a, 0, sizeof(a));
        for(i=1; i<=n; ++i)
            a[i] = i;
        t = 0;
        left = 0, right = n;
        while(t < n)
        { 
            if( t )
                printf(",");
            i = j = 0;
            while(!a[left]) 
            { 
                left++;
                left = left%n ? left%n : n;
            }
            while(!a[right])
            { 
                right--; 
                right = right <= 0 ? n : right;
            }
            while(i<k)
            { 
                if(a[(left = left%n? left%n: n)])
                    i++;
                if(i >= k)
                    continue; 
                left = left%n + 1;
            }
            while(j<m)
            { 
                if(a[(right = right%n ? right%n : n)])
                    j++;
                if(j >= m)
                    continue;
                right = right - 1 ? right - 1 : n;
            }
            printf("%3d", a[left]);
            a[left] = 0; 
            t++;
            if(left != right)
            { 
                printf("%3d", a[right]);
                a[right] = 0; 
                t++;
            } 
        }
        printf("\n");
    }
    return 0;
}