标签:nyoj714
The magician shuffles a small pack of cards, holds it face down and performs the following procedure:
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.
2
4
5
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;
}
NYOJ714 Card Trick 【队列模拟】,布布扣,bubuko.com
标签:nyoj714
原文地址:http://blog.csdn.net/chang_mu/article/details/25558687