UVA133 The Dole Queue
2014-07-20 20:40
344 查看
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;
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 = 1; if(num2 == n + 1) p[1] = 1; p[num1] = 1; p[num2] = 1; } memset(p , 0 ,sizeof(p)); } return 0; }
相关文章推荐
- UVa-133-The Dole Queue
- UVa-133-The Dole Queue
- 数据结构 uva 133 - The Dole Queue
- The Dole Queue UVA - 133
- uva133-The Dole Queue
- UVA-133 The Dole Queue
- uva 133 The Dole Queue(循环队列)
- 救济金发放The Dole QueueUVA 133
- UVa 133 - The Dole Queue 数据结构专题
- UVA - 133 The Dole Queue
- The Dole Queue——Uva_133
- uva 133 The Dole Queue
- UVa-133 The Dole Queue
- 救济金发放 (The Dole Queue UVa 133)
- uva-133 The Dole Queue
- UVA - 133 The Dole Queue
- Uva 133 - The Dole Queue
- UVA 133 - The Dole Queue
- UVa133 - The Dole Queue
- 算法竞赛入门经典(紫书)第四章—— The Dole Queue UVA-133