【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 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 railing 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n开始顺时针数。在每一轮中，官员A数k个就停下来，官员B数m个就停下来（注意有可能两个官员停在同一个人上）。接下来被官员选中的人（1个或者2个）离开队伍。输入n，k，m输出每轮里被选中的人的编号（如果有两个人，先输出被A选中的）。例如，n=10，k=4，m=3，输出为48,95,31,26,10,7。注意：输出的每个数应当恰好占3列。

思路：
用一个大小为0的数组表示人站成的圈。为了避免人走之后移动数组元素，用0表示离开队伍的人，数数时跳过即可。

AC代码：

#include <iostream>
#include <cstdio>
int run(int x, int y, int z); // x:数的个数 y:从y开始数 z:数的方向
int arr[22];
int n, k, m;
using namespace std;
int main()
{while (cin >> n >> k >> m){if (n == 0 && k == 0 && m == 0){return 0;}for (int i = 0; i <= n; i++){arr[i] = i;}int num = n, t1 = n, t2 = 1;while (num){t1 = run(k, t1, 1);t2 = run(m, t2, -1);printf("%3d", arr[t1]);num--;if (t1 != t2){printf("%3d", arr[t2]);num--;}arr[t1] = arr[t2] = 0;if (num){cout << ",";}}cout << endl;}return 0;
}int run(int x, int y, int z)
{while (x--){do{y = (y + z + n -1) % n + 1;} while (arr[y] == 0);}return y;
}

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【UVA】 133 --- The Dole Queue

【UVA】 133 --- The Dole Queue

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