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NYOJ 714 Card Trick
Card Trick
时间限制:1000 ms | 内存限制:65535 KB
难度: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.
- 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
2
4
5
- 样例输出
2 1 4 3
3 1 4 5 2
模拟!
AC码:
#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; }
NYOJ 714 Card Trick
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