The Dole Queue（子过程设计）（UVa 133）

The Dole Queue

Time Limit: 3000ms

Memory Limit: 131072KB
This problem will be judged on UVA.
Original ID: 133

64-bit integer IO format: %lld     
Java class name: Main

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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 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.
代码分析：go（）函数把一个走步的复杂功能从main中分离出来，增加了代码可读性。
#include<stdio.h>
#define MAXN 25

int n, k, m, a[MAXN];

int go(int p, int d, int k) //起点坐标p，方向d（1逆时针，-1顺时针），步长k
{
while(k)
{

while(a[(p = (p+d+n-1)%n+1)] == 0); //p = (p+d+n-1)%n+1保证p永远不会越界
k--;
}
return p;
}

int main()
{
int p1, p2, left;
while(scanf("%d %d %d", &n, &k, &m) == 3)
{
if(n==0 && k==0 && m==0) break;

p1 = n; //因为开始的起点算一步，所以从起点前一步开始算起
p2 = 1; //同上
left = n;
for(int i=1; i<=n; i++) a[i] = 1; //编号为1~n

while(left)
{
p1 = go(p1, 1, k);
p2 = go(p2, -1, m);

printf("%3d", p1);left--;a[p1]=0;
if(p1 != p2) {printf("%3d", p2);left--;a[p2]=0;}
if(left) printf(",");

}

printf("\n");

}
return 0;
}