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 counter-clockwise 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 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 3
0 0 0
Sample output
4 8, 9 5, 3 1, 2 6, 10, 7
where represents a space.
這個題的意思是給你一個長度為N的循環隊列,一個人從1號開始逆時針開始數數,第K個出列,一個人從第N個人開始順時針數數,第M個出列,選到的兩個人要同時出列(以不影響另一個人數數),選到同一個人只出隊一次。
解題思想:用數組模擬,arr[i]等於0時表示第I個人出列。
#include <iostream> #include<deque> #include<algorithm> #include<cstdio> using namespace std; int main() { int n,k,m,size; int arr[20]; while(cin>>n>>k>>m) { if(n==0&&k==0&&m==0)break; int p=0,q=n-1; size=n; for(int i=0; i<n; i++) arr[i]=i+1; while(size>0) { for(int i=1; i<k; i++) { p=(p+1)%n; if(!arr[p])i--; } for(int i=1; i<m; i++) { q=(q-1+n)%n; if(!arr[q])i--; } if(arr[p]) { printf("%3d",arr[p]); arr[p]=0; size--; } if(arr[q]) { printf("%3d",arr[q]); arr[q]=0; size--; } for(int i=0; i<n; i++) { p=(p+1)%n; if(arr[p])break; } for(int i=0; i<n; i++) { q=(q-1+n)%n; if(arr[q])break; } if(size)printf(","); else printf("\n"); } } return 0; }