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
     
      #define maxn 25int 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("\n");    }    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
     
      #include
      
       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;}