POJ 3032 Card Trick

Card Trick

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3224		Accepted: 2354

Description

The magician shuffles a small pack of cards, holds it face down and performs the following procedure:

1. 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.
2. 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.
3. Three cards are moved one at a time…
4. 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

Source

Nordic 2006

 

 

题意：给你n张牌，让你变一个魔术：第1次把上面的1张牌放到底部，然后最上面的牌就是1，然后拿走1。第2次把上面的2张牌依次放到底部，然后最上面的牌就是2，然后拿走2....重复这个过程，直到所有的牌都被拿走。问一开始的牌应该从上到下怎么放，才能完成这个魔术。

算法：逆向思维，从后向前模拟。不要“下盲棋”，拿几张牌试试就明白了。。。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

int n;
queue<int> q;

void output(){  // 递归输出q里面的内容
    int tmp=q.front();
    q.pop();
    while(!q.empty())
        output();
    printf("%d ",tmp);
}

int main(){

    //freopen("input.txt","r",stdin);

    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        int cnt;
        while(n>0){
            q.push(n);
            cnt=n;
            n--;
            while(cnt--){    // 由后向前翻cnt次牌
                int tmp=q.front();
                q.pop();
                q.push(tmp);
            }
        }
        output();
        printf("\n");
    }
    return 0;
}