L - The Dole Queue

青旅半醒 2022-10-01 04:51 3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;
    }