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题面：

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 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 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

题目大意：

一共有n个人，坐成一个环。以1-n逆时针编号。有两个人A,B。A从1开始数k个人，B从n开始数m个人。被数到的人退场。然后继续数，直到所有人都退场。

输出退场次序，若A与B数的人编号不同，A数的先输出。

大致思路：

开一个数组保存编号。退场的编号为0，数到0的时候跳过。记得对A取膜，B小于零时变为n。

代码：

#include
   
    using namespace std;bool isempty(int *s,int n)//判断是否全部退场{    for(int i=1;i<=n;++i)        if(s[i]!=0)            return true;    return false;}int main(){    int n,k,m;    int s[21];    while(cin>>n>>k>>m&&(n||k||m))    {        int a=1,b=n,cnt=1;        for(int i=1;i<=n;++i)            s[i]=i;        while(isempty(s,n))        {            if(cnt!=1)                printf(",");            int ca=1,cb=1;            while(ca!=k||cb!=m)            {                if(ca
    
     n)                        a%=n;                    if(s[a]!=0)                        ca++;                }                if(cb
     
      n)                        a=1;                }                while(s[b]==0)                {                    --b;                    if(b<1)                        b=n;                }            }            cnt++;        }        printf("\n");    }    return 0;}
     
    
   

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