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uva 133 - The Dole Queue
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
1 #include<iostream> 2 #include<string.h> 3 #include<stdio.h> 4 #include<ctype.h> 5 #include<algorithm> 6 #include<stack> 7 #include<queue> 8 #include<set> 9 #include<math.h> 10 #include<vector> 11 #include<map> 12 #include<deque> 13 #include<list> 14 using namespace std; 15 #define maxn 2516 int n,k,m,a[maxn];17 int go(int p,int d,int t)//构建走动函数,跳过数值为0的位置 18 {19 while(t--)20 {21 do22 {23 p=(p+d+n-1)%n+1;24 }25 while(a[p]==0);26 }27 return p;28 } 29 int main()30 {31 while(scanf("%d%d%d",&n,&k,&m)==3&&n)32 {33 for(int i=1;i<=n;i++)34 a[i]=i;35 int left =n;36 int p1=n,p2=1;//初始化移动位置,逆时针用p1,顺时针用p237 while(left)38 {39 p1=go(p1,1,k);40 p2=go(p2,-1,m);41 printf("%d",p1);42 left--;43 if(p1!=p2)44 {45 printf(" %d",p2);46 left--;47 }48 if(left)49 printf(",");50 a[p1]=a[p2]=0;;51 } 52 printf("\n");53 }54 return 0;55 }