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NYOJ 714 Card Trick

發布時間:2025/3/16 编程问答 26 豆豆
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Card Trick

時間限制:1000?ms ?|? 內存限制:65535?KB 難度:3 描述

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.

    輸入
    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
    輸出
    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…
    樣例輸入
    2
    4
    5
    樣例輸出
    2 1 4 3
    3 1 4 5 2
    模擬!
    AC碼:
    #include<stdio.h> int main() {int T,n,i,k,step;int visit[15],num[15];scanf("%d",&T);while(T--){scanf("%d",&n);for(i=0;i<=n;i++)visit[i]=0;k=1;i=1;step=0;while(k<=n){if(visit[i]==0){if(step==k){num[i]=k;k++;step=-1;visit[i]=1;}step++;}i++;if(i>n)i=1;}for(i=1;i<=n;i++)printf("%d ",num[i]);printf("\n");}return 0; }

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