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(" "); else printf(","); } } }