水题：UVa133-The Dole Queue

The Dole Queue

Time limit 3000 ms

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 counterclockwise 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

Writeaprogramthatwillsuccessivelyreadin(inthatorder)thethreenumbers(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 counterclockwise official first. Separate successive pairs (or singletons) by commas (but there should not be a trailing comma).
Note: The symbol ⊔ in the Sample Output below represents a space.

Sample Input

10 4 3 0 0 0

Sample Output

␣␣4␣␣8,␣␣9␣␣5,␣␣3␣␣1,␣␣2␣␣6,␣10,␣␣7

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解题心得：

1. 紫书上的例题，很简单可以直接模拟整个过程就行了，但是要注意输出的问题（”%3d”）。在一个功能的模块需要多次使用的时候可以直接写成函数，减少代码长度，不容易出错。

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5;
bool cir[maxn];
int r,l,Len;

int _find(int pos,int len,int dir)//dir代表方向，1代表向右，0代表向左，pos是当前位置，len代表要走多少步
{
    while(len)
    {
        if(dir == 1)
            pos++;
        else
            pos--;
        if(pos > Len)
            pos = 1;
        if(pos < 1)
            pos = Len;
        if(!cir[pos])
            len--;
    }
    return pos;
}

int main()
{
    int len,m,k;
    while(scanf("%d%d%d",&len,&m,&k) && len+m+k)
    {
        Len = len;
        memset(cir,false,sizeof(cir));
        bool flag = false;
        r = 0,l = Len+1;
        while(len)
        {
            r = _find(r,m,1);
            l = _find(l,k,0);
            cir[r] = cir[l] = true;//true代表已经退出了的
            if(r == l)
            {
                if(!flag)//flag用来控制,的输出
                    printf("%3d",r);
                else
                    printf(",%3d",r);
                len--;
            }
            else
            {
                if(!flag)
                    printf("%3d%3d",r,l);
                else
                    printf(",%3d%3d",r,l);
                len -= 2;
            }
            flag = true;
        }
        printf("
");
    }
    return 0;
}