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
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5
样例输出 -
2 1 4 3
-
3 1 4 5 2
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比较简单,n值最大才13.
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#include<stdio.h> int a[15]; int main() { int t,n,i,j,m,k; int cishu; scanf("%d",&t); while(t--) { scanf("%d",&n); //最大的两个数顺序先找出 if((n-1)&1) { a[n]=n-1; a[n-1]=n; } else { a[n]=n; a[n-1]=n-1; } //找剩余的 for(i=n-2;i>=1;i--) { a[i]=i; cishu=i; while(cishu--) { k=a[n];//保存最后一位的值 for(j=n;j>=i-1;j--) { a[j]=a[j-1]; } a[i]=k;//赋值第一位 } } for(i=1;i<=n;i++) { if(i>1) printf(" "); printf("%d",a[i]); } printf("\n"); } return 0; }