nyoj 714 第六届河南省大学生程序设计竞赛F

Card Trick

时间限制：1000 ms  |  内存限制：65535 KB

描述

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.

输入

On the first line of the input is a single positive integer k, telling the number of test cases to follow. 1 ≤ k ≤ 10 Each case consists of one line containing the integer n. 1 ≤ n ≤ 13

输出

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…

样例输入

245

样例输出

2 1 4 33 1 4 5 2

来源

第六届河南省程序设计大赛

上传者

ACM_赵铭浩

第六届的题好难-.-前面的都没思路-.-.-.-

题意：1-n，n张牌，魔法师操作n次，第i次从上面移动i个牌到下面，然后把最上面的那个掀开..

求开始时牌的顺序使得牌掀开的顺序是1-n..

模拟-.-

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n,shu[15];bool fafe[15];int main(){int t;scanf("%d",&t);while (t--){scanf("%d",&n);int lp=1;memset(fafe,true,sizeof(fafe));for (int i=1;i<=n;i++){int ge=i;int j=0;while (j<ge){if (fafe[lp]){j++;    lp++;}else lp++;if (lp==n+1) lp=1;}if (!fafe[lp]){while (1){    lp++;if (lp==n+1) lp=1;    if (fafe[lp]) break;    }}shu[lp]=i;fafe[lp]=false;}printf("%d",shu[1]);for (int i=2;i<=n;i++)printf(" %d",shu[i]);printf("\n");}return 0;}