UVA133 The Dole Queue

最新推荐文章于 2019-07-29 14:20:35 发布

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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 3
0 0 0

Sample output

4 8, 9 5, 3 1, 2 6, 10, 7

where represents a space.

题目意思就是ｎ个人围一圈，然后两个官员，一个从１开始往大的方向数，数到第ｋ个就选中。。一个是从ｎ开始往小的方向数，数到第ｍ个选中，数的时候被选中的要跳过。

注意是两人同时数完一次，然后这两人才被标记，并且第一个官员先输出，第二个官员后输出，如果两人选中同一个，则只输出一个。直到全选中完。。

用数组模拟即可；

ＡＣ代码如下；

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;

int n,k,m;
int count;
int num1,num2;
int p[30] = {0};
int judge(int num) {
	for (int l = 0 ; l < k ;) {
	
		if(p[num++] == 0)
			l++;
		if (num > n)
			num = 1;
	}
	return num - 1;
}

int judge2(int num) {
	for (int l = 0; l < m;) {
	
		if(p[num--] == 0)
			l++;
		if(num < 1)
			num = n;
	}
	return num + 1;


}

int main () {
	while (cin >> n >> k >> m) {
		if (n == 1) {
		
			printf("  1\n");
			continue;

		}
		count = 0;
		if(n + k + m == 0)
			break;
		num1 = 0;
		num2 = n + 1;
		while (count < n) {
			num1 = judge (num1 + 1);
			num2 = judge2 (num2 - 1);
			if (num1 == 0)
				printf("%3d",n);
			else
				printf("%3d",num1);
			count++;
			if((num2 != num1) && !(num1 == 0 && num2 == n) && !(num1 == 1&&num2 == n + 1)) {
				if (num2 == n + 1)
					printf("%3d",1);
				else
					printf("%3d",num2);
				count++;
			}
			if (count != n)
				printf(",");
			else 
				printf("\n");
			if(num1 == 0)
				p[n] = 1;
			if(num2 == n + 1)
				p[1] = 1;
			p[num1] = 1;
			p[num2] = 1;
		}
		memset(p , 0 ,sizeof(p));
	}
	return 0;
}