1、http://poj.org/problem?id=3032
2、
Card Trick
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 3231 | Accepted: 2359 |
Description
The magician shuffles a small pack of cards, holds it face down and performs the following procedure:
- The top card is moved to the bottom of the pack. The new top card is dealt face up onto the table. It is the Ace of Spades.
- Two cards are moved one at a time from the top to the bottom. The next card is dealt face up onto the table. It is the Two of Spades.
- Three cards are moved one at a time…
- This goes on until the nth and last card turns out to be the n of Spades.
This impressive trick works if the magician knows how to arrange the cards beforehand (and knows how to give a false shuffle). Your program has to determine the initial order of the cards for a given number of cards, 1 ≤ n ≤ 13.
Input
On the first line of the input is a single positive integer, telling the number of test cases to follow. Each case consists of one line containing the integer n.
Output
For each test case, output a line with the correct permutation of the values 1 to n, space separated. The first number showing the top card of the pack, etc…
Sample Input
2 4 5
Sample Output
2 1 4 3 3 1 4 5 2
3、代码：
#include<stdio.h> #include<deque> #include<algorithm> using namespace std; int main() {int t,n;scanf("%d",&t);while(t--){scanf("%d",&n);deque<int> q;q.push_back(n);for(int i=n-1;i>=1;i--){q.push_back(i);for(int j=1;j<=i;j++){int temp=q.front();q.push_back(temp);q.pop_front();}}while(!q.empty()){printf("%d ",q.back());q.pop_back();}printf("\n");}return 0; }