UVA - 133 The Dole Queue
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Description
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.
约瑟夫问题。
用数组a标记,0既为离开队伍的,数数时跳过。
#include<iostream>#include<iomanip>using namespace std;const int MAXN = 25;int n, a[MAXN];int go(int p, int d, int t) //p是报数开始的前一个位置,第一次报数时即n(正数)和1(倒数)。{ //抽象出步长d的参数,把正数(1)与倒数(-1)两个go统一。t是步数while (t--){do{ p = (p + d + n -1 ) % n + 1; } //一开始走到报数的第一个数。while (a[p] == 0); //走到下一个非0数字。}return p; //返回要离开的数}int main(){int k, m;while (cin >> n >> k >> m&&n){for (int i = 1; i <= n; i++)a[i] = i;int left = n; //left记录剩下的有多少个。int p1 = n, p2 = 1; //注意,如上所说的。while (left){p1 = go(p1, 1, k);p2 = go(p2, -1, m);if (p1 == p2) //区分两种情况。{cout<<setw(3) << p1;a[p1] =a[p2]= 0;left--;}else{cout << setw(3) << p1;cout << setw(3) << p2;a[p1] = a[p2] = 0;left--;left--;}if (left)cout << ',';}cout << endl;}}
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