UVA

/*  这题用到了队列的思想    BTW，查题解时，看到两个大佬的代码真是简洁，于是果断抛弃了自己写的，用他们的思路重新做了一遍    参考博客：  http://www.cnblogs.com/AlgoWing/archive/2013/03/04/3189616.html  http://blog.csdn.net/mobius_strip/article/details/45157587    另附C++中对于队列的讲解blog：  http://www.cnblogs.com/xuning/p/3321733.html  (注意，队列的size的返回值并不是int型，所以使用前应先强制转换，老生常谈了...)*/

//法一:#include <iostream>using namespace std;const int maxn = 100000;int Q[maxn];int main(){int n;while (cin >> n && n){for (int i = 1; i <= n; i++) Q[i] = i;int head = 1, tail = n, first = 1;cout << "Discarded cards:"; //注意冒号后无空行 while (head < tail){if (!first) cout << ",";cout << " " << Q[head];head++;Q[++tail] = Q[head++];first = 0; //除了第一次不用输出逗号，其他时候都要 }cout << endl << "Remaining card: " << Q[head] << endl;}return 0;}

//法二：#include <iostream>#include <queue>using namespace std;queue<int> Q;int n;int main(){while (cin >> n && n){for (int i = 1; i <= n; i++) Q.push(i);cout << "Discarded cards:";while ((int)Q.size() > 1){if ((int)Q.size() > 2){cout << " " << Q.front() << ",";Q.pop();Q.push(Q.front());Q.pop();}else{cout << " " << Q.front();Q.pop();}}cout << endl << "Remaining card: " << Q.front() << endl;Q.pop();}return 0;}