首页 > 代码库 > NYOJ714 Card Trick 【队列模拟】
NYOJ714 Card Trick 【队列模拟】
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
/* **题意是给定一个1到n的某种初始序列,然后将列首数放到序列的末尾(操作一次),将新的队首元素放到 **桌面,再将队首的两个数放到末尾(操作两次),新树放到桌面,以此类推。最终得到的数列是1到n的递 **增序列。求初始序列。 **这是一道模拟题,可以逆推:由于最终序列一定是1,2,3...n,所以最后一个摆在桌面上的数一定是n, **且n被“操作”了n次,倒数第二个被放到桌面的数一定是n-1且n-1一定和n一起被“操作”了n-1次,以此类推。 **可以倒着模拟,即先将最终数列放到数组里,再从第n个数开始模拟,即将第n个数向前“操作”n次,实际上 **位置没变,然后将第n-1个数加入到“操作”中且与第n个数一起向前“操作”n-1次,以此类推,最终得到的数 **列即是初始序列。 */ #include <stdio.h> #include <string.h> int a[15]; void move(int bot, int top, int n){ while(n--){ int t = a[top], tt; for(int i = bot; i <= top; ++i){ tt = a[i]; a[i] = t; t = tt; } } } int main(){ int t, n, bot, top; scanf("%d", &t); while(t--){ scanf("%d", &n); for(int i = 1; i <= n; ++i) a[i] = i; top = bot = n; for(int i = n; i >= 1; --i){ move(bot, top, i % (top - bot + 1)); --bot; } for(int i = 1; i <= n; ++i) printf("%d ", a[i]); printf("\n"); } return 0; }
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