Uva 133 The Dole Queue 双向约瑟夫环
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.
题目大意：
(1) N个人围成一圈。随便指定一个位置为1，然后依次逆时针标记为1,2,...,N。
(2) 一个人从1开始数到 k，这个人出列，为A[i]，另一个人同时从 N数 m个数，即 N-m出列，记为B[i]。
(3) 因为是同时计数的,所以可能出现都标记到同一个人,这时A[i] = B[i]。只输出一个即可。
否则输出时都是成对输出，然后输出一个 "," 当然末尾没有"," (当然 如果A[i] = B[i]就只输出一个数了)
注意：让两个人都找到A[i]，B[i] 后再标记A[i]，B[i]已经访问，不然无法得到正确输出。
#include <iostream>
#include <string>
#include <stdio.h>
using namespace std;
string str;
void init(int n) {
str = "";
char ch;
for(int i = 1;i <= n;i++) {
ch = 1;
str += ch;
}
}
void solve(int n,int l,int r) {
int cut_l = 0;
int cut_r = 0;
int cut = 0;
int i=0,j=n-1;
int flag1,flag2;
while(cut < n) {
while(cut_l < l ) {
if(str[i] != 0)
cut_l++;
if(cut_l == l)
flag1=i;
if(i < n-1)
i++;
else
i=0;
}
while(cut_r < r) {
if(str[j] != 0)
cut_r++;
if(cut_r == r)
flag2 = j;
if(j > 0)
j--;
else
j=n-1;
}
if(flag1 == flag2) {
str[flag1] = str[flag2] =0;
cut++;
printf("%3d",flag1+1);
}
else{
str[flag1] = str[flag2] = 0;
cut += 2;
printf("%3d%3d",flag1+1,flag2+1);
}
if(cut != n) {
cout<<',';
}
cut_l=0,cut_r=0;
}
}
int main() {
int n,l,r;
while(cin >> n >> l >> r && (n || l || r)) {
init(n);
solve(n,l,r);
cout<<endl;
}
return 0;
}