The Dole Queue UVA - 133

标签：ios   log   out   cout   cin   ide   i++   ora   scanf   

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 counterclockwise 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 Writeaprogramthatwillsuccessivelyreadin(inthatorder)thethreenumbers(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 counterclockwise official first. Separate successive pairs (or singletons) by commas (but there should not be a trailing comma). Note: The symbol ? in the Sample Output below represents a space. Sample Input 10 4 3 0 0 0 Sample Output ??4??8,??9??5,??3??1,??2??6,?10,??7

题意：n(n<20)个人站成一圈，逆时针编号为1~n。有两个官员，A从1开始逆时针数，B从顺时针开始数。在每一轮中，A数K个就停下来，B数m个就停下来（可能停在同一个人身上）。接下来被官员选中的人（1个或2个）离开队伍。先输出被A选中的！

注意：p=(p+d+n-1)%n+1;可以自己模拟几遍；

　　   do{ }while在此题很应景；

　　　归纳出相似代码的特点并用函数简化，例如:https://vjudge.net/problem/18191/origin(The Blocks Problem UVA - 101 )

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 using namespace std;
 6 
 7 int num[30];
 8 int n,x,y;
 9 
10 int go(int p,int d,int t){
11     while(t--){    
12         do{
13             p=(p+d+n-1)%n+1;
14         }while(num[p]==0);
15     }
16     return p;
17 }
18 
19 int main()
20 {    
21     while(~scanf("%d%d%d",&n,&x,&y)){
22         if(n==0||x==0||y==0) break;
23         
24         for(int i=1;i<=n;i++) num[i]=i;
25         
26         int left=n;
27         int p=n,q=1;
28         while(left){
29             p=go(p,1, x);
30             q=go(q,-1,y);
31             printf("%3d",p);
32             left--;
33             if(p!=q){
34                 printf("%3d",q);
35                 left--;
36             }
37             num[p]=0,num[q]=0;
38             if(left) printf(",");
39         }
40         cout<<endl;
41     }
42     return 0;    
43 }

The Dole Queue UVA - 133

标签：ios   log   out   cout   cin   ide   i++   ora   scanf   

原文地址：http://www.cnblogs.com/zgglj-com/p/7236978.html