The Dole Queue UVA

刘佳汝书上的例题，82页4-3，没什么难度，他说的很详细了，我稍微注释了一下。

#include<stdio.h>
#define maxn 25
int n, k, m, a[maxn];

int go(int p, int d, int t)//用d的+-1取值来代表向前走和向后走
{
    while (t--)
    {
        do { p = (p + d + n - 1) % n + 1; } while (a[p] == 0);//走到下一个非零数字其中-1是为了平衡d为+1的情况，最后+1是为了平衡d为-1的情况
    }
    return p;
}


int main()
{
    while (scanf("%d%d%d", &n, &k, &m) == 3 && n)
    {
        for (int i = 1; i <= n; i++) a[i] = i;
        int left = n; //剩余人数
        int p1 = n, p2 = 1;
        while (left)
        {
            p1 = go(p1, 1, k);
            p2 = go(p2, -1, m);
            printf("%3d", p1); left--;
            if (p2 != p1) { printf("%3d", p2); left--; }
            a[p1] = a[p2] = 0;
            if (left) printf(",");
        }
        printf("
");
    }
    return 0;
}

下边的今天自己写的，说实话他那个卡范围的表达式是真的难想，退而求其次了。

还有，%3d可真好用！

#include<iostream>
#include<iomanip>

using namespace std;
int Q[25];
int k, m, n;

int go(int t, int x, int i)
{
    while (x--)
    {
        do
        {
            t += i;
            if (t <= 0) t = n;
            if (t >= n + 1)t = 1;
        } while (!Q[t]);
    }
    return t;
}

int main()
{
    cin.tie(0);
    cin.sync_with_stdio(false);
    //memset(Q, 0, sizeof(Q));
    while (cin >> n >> k >> m)
    {
        if (!(k || m || n))
        {
            break;
        }
        int _cnt = n;
        //初始化队列
        for (int i = 0; i <= n; i++)
        {
            Q[i] = i;
        }
        //发放救济金
        int a = 0, b = 0;
        while (_cnt)
        {
            a = go(a, k, 1);
            b = go(b, m, -1);
            if (a == b)
            {
                Q[a] = 0;
                cout << setiosflags(ios::fixed) << setiosflags(ios::right);
                cout << setw(3) << a;//格式化输出
                _cnt--;
            }
            else
            {
                Q[a] = 0; Q[b] = 0;
                cout << setw(3) << a << setw(3) << b;
                
                _cnt -= 2;
            }
            if(_cnt!=0)
                cout<< ",";
        }
        cout << endl;
    }
    return 0;
}