Uva 133 The Dole Queue 双向约瑟夫环
2014-07-21 09:26
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C - The Dole Queue
Time Limit:0MS Memory Limit:0KB 64bit IO Format:%lld & %llu
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.
end of data will be signalled by three zeroes (0 0 0).
pairs (or singletons) by commas (but there should not be a trailing comma).
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。
Time Limit:0MS Memory Limit:0KB 64bit IO Format:%lld & %llu
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 theend 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 successivepairs (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; }
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