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UVA – 133

编程语言 Code_Diary 150℃ 0评论

In a serious attempt to downsize (reduce) the dole queue, The New National Green Labour RhinocerosParty has decided on the following strategy. Every day all dole applicants will be placed in a largecircle, facing inwards. Someone is arbitrarily chosen as
number 1, and the rest are numbered counterclockwiseup 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 againat 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 alreadyselected 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 ofthree 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. Eachnumber should be in a field of 3 characters. For pairs of numbers list the person chosen by the counterclockwiseofficial first. Separate successive pairs
(or singletons) by commas (but there should not be atrailing comma).Note: The symbol ⊔ in the Sample Output below represents a space.

Sample Input

10 4 30 0 0

Sample Output

␣␣4␣␣8,␣␣9␣␣5,␣␣3␣␣1,␣␣2␣␣6,␣10,␣␣7

(␣␣这是两个空格)

#include
#include
using namespace std;
 int main()
 {
  int N, k, m;
  while(~scanf("%d %d %d", &N, &k, &m))
  {
   if(N == 0||k == 0||m==0)break;
   int f[30] = {0}, num = 0, km = 0, mm = 0, i = 1, j = N, flag = N;
   while(num != N)
   {
    while(1)
    {
     if(i > N)
      i = 1;
     if(f[i] != 1)
      ++km;
     if(km == k)
     {
      km = 0;
      ++num;
      break;
     }
     ++i;
    }
    while(1)
    {
     if(j < 1)
      j = N;
     if(f[j] != 1)
      ++mm;
     if(mm == m)
     {
      mm = 0;
      f[j] = 1;
     f[i] = 1;
     if(i != j)
       ++num;
      break;
     }
     --j;
    }
    if(i == j)
    {
     printf("%3d", i);
     --flag;
    }
    else if(i != j)
    {
     printf("%3d%3d",i,j);
     flag -= 2;
    }
    if(flag)
     printf(",");
   }
   printf("\n");
 }
}





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