Uva 133详解

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.

解题思路：

模拟游戏进行的过程，首先输入三个数，分别是人数n、逆数k、顺数m；

分配数组内存，用于记录每个序号的人在与不在，分别用1和0表示；

初始化数组

循环进行每轮的筛选，num记录被筛选出的人数

对于官员A来说，顺数k个a[i] = 1的，将第k个人的序号用x记录下来；

对于官员B来说，逆数m个a[i]= 1的，将第m个人的序号用y记录下来；

再判断x和y的序号是否相等来决定输出的内容，以及num的变化

---

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
int main()
{
    int n, k, m;
    while(cin >> n >> k >> m && n != 0)
    {
        int *a = (int *)malloc((n + 1) * sizeof(int));
        for(int i = 0; i < n + 1; i++)
           a[i] = 1;
        int x, y;
        int p1 = 1, p2 = n;
        for(int num = 0; num < n; )
        {
            for(int i = 0; i < k; )
            {
                if(a[p1] == 1)
                {
                    if(i == k - 1)
                       x = p1;
                    i++;
                }
                p1++;
                if(p1 == n + 1)
                   p1 = 1;
            }
            for(int i = 0; i < m; )
            {
                if(a[p2] == 1)
                {
                    if(i == m - 1)
                       y = p2;
                    i++;
                }
                p2--;
                if(p2 == 0)
                  p2 = n;
            }
            printf("%3d", x);
            a[x] = 0; num++;
            if(x != y)
            {
                printf("%3d", y);
                a[y] = 0;
                num++;
            }
            if(num == n)
               printf("\n");
            else
               printf(",");
        }
    }
}