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<iostream> #include<cstdio> 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("=", i); --flag; } else if(i != j) { printf("==",i,j); flag -= 2; } if(flag) printf(","); } printf("\n"); } }