UVa 133 - The Dole Queue

 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 30 0 0

Sample output

 4  8,  9  5,  3  1,  2  6,  10,  7

where  represents a space.

#include <cstdio>#include <cstdlib>struct Person{struct Person *front;int value;struct Person *next;};Person *top, *end;int n, k, m;int nCase=0;void solve(){if(nCase)printf("\n");Person *left=top, *right=end;Person *left_last = left, *right_last=right;while(1){for(int i=2; i <= k; i++){left_last= left;left = left->next;}for(int j=2; j <= m; j++){right_last = right;right=right->front;}if(left==right){if(n==1){printf("%3d", left->value);break;}else{printf("%3d,", left->value);right_last->front = left_last;left_last->next = right_last;left = left->next;right = right->front;n--;}}else{printf("%3d",left->value);left_last->next = left->next;left->next->front = left_last;printf("%3d",right->value);right->front->next = right_last;right_last->front = right->front;n -=2;if(n==0) break;else printf(",");Person *l=left, *r=right;l = l->next;r = r->front;if(l==right)l = l->next;if(r==left)r = r->front;left = l;right = r;}}}void init(){Person *p = (Person *)malloc(sizeof(Person));Person *pre=NULL;top = p; pre = p;p->front = NULL;p->next=NULL;p->value = 1;for(int i=2; i <= n; i++){Person *person = (Person *)malloc(sizeof(Person));pre->next = person;person->front = pre;person->next = NULL;person->value = i;pre = pre->next;}pre->next = p;p->front = pre;end = pre;}int main(){//freopen("in.txt","r",stdin);while(scanf("%d %d %d", &n, &k, &m)==3 &&n!=0 && k!=0 && m!= 0){init();solve();nCase++;}return 0;}

以上是我的代码，纯用双向链表来模拟实现，既复杂又没有AC......郁闷的是我找了一个晚上都不知道错在哪里.....伤心了之后，去网上看到别人写的这些代码，我当时就跪了......

#include <cstdio>  #include <cstring>  const int maxn = 22;    int N;  bool d[maxn];    int cw(int pos, int k) {      int cnt = 0;      for ( ; cnt != k; pos++) {          if (pos == N) pos = 0;          if (d[pos]) cnt++;      }      return --pos;  }    int w(int pos, int m) {      int cnt = 0;      for (; cnt != m; pos--) {          if (pos == -1) pos = N - 1;          if (d[pos]) cnt++;      }      return ++pos;  }  int main() {      int k, m;      while (scanf("%d%d%d", &N, &k, &m) && N && k && m) {          memset(d, 1, N);          int tot = N, flag = 0, a = 0, b = N - 1;          while (tot) {              a = cw(a, k), b = w(b, m);              tot--;              if (flag) printf(",");              d[a] = d[b] = false;              flag = 1;              if (a == b)                  printf("%3d", a + 1);              else {                  printf("%3d%3d", a + 1, b + 1);                  tot--;              }          }//while find out all          printf("\n");      }//while      return 0;  }  

数组代替链表，布尔变量标记是否访问过............

尼玛，不活了！！！