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  • Uva 133详解

    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

    tex2html_wrap_inline34 4 tex2html_wrap_inline34 8, tex2html_wrap_inline34 9 tex2html_wrap_inline34 5, tex2html_wrap_inline34 3 tex2html_wrap_inline34 1, tex2html_wrap_inline34 2 tex2html_wrap_inline34 6, tex2html_wrap_inline50 10, tex2html_wrap_inline34 7

    where tex2html_wrap_inline50 represents a space.

    解题思路:

    模拟游戏进行的过程,首先输入三个数,分别是人数n、逆数k、顺数m;

    分配数组内存,用于记录每个序号的人在与不在,分别用1和0表示;

    初始化数组

    循环进行每轮的筛选,num记录被筛选出的人数

    对于官员A来说,顺数k个a[i] = 1的,将第k个人的序号用x记录下来;

    对于官员B来说,逆数m个a[i]= 1的,将第m个人的序号用y记录下来;

    再判断x和y的序号是否相等来决定输出的内容,以及num的变化


    #include<iostream>
    #include<algorithm>
    #include<stdio.h>
    using namespace std;
    int main()
    {
        int n, k, m;
        while(cin >> n >> k >> m && n != 0)
        {
            int *a = (int *)malloc((n + 1) * sizeof(int));
            for(int i = 0; i < n + 1; i++)
               a[i] = 1;
            int x, y;
            int p1 = 1, p2 = n;
            for(int num = 0; num < n; )
            {
                for(int i = 0; i < k; )
                {
                    if(a[p1] == 1)
                    {
                        if(i == k - 1)
                           x = p1;
                        i++;
                    }
                    p1++;
                    if(p1 == n + 1)
                       p1 = 1;
                }
                for(int i = 0; i < m; )
                {
                    if(a[p2] == 1)
                    {
                        if(i == m - 1)
                           y = p2;
                        i++;
                    }
                    p2--;
                    if(p2 == 0)
                      p2 = n;
                }
                printf("%3d", x);
                a[x] = 0; num++;
                if(x != y)
                {
                    printf("%3d", y);
                    a[y] = 0;
                    num++;
                }
                if(num == n)
                   printf("
    ");
                else
                   printf(",");
            }
        }
    } 
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  • 原文地址:https://www.cnblogs.com/denghui666/p/8387887.html
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