707
NYOJ714-Card Trick
Card Trick
時間限製：1000 ms | 內存限製：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
245
樣例輸出
2 1 4 33 1 4 5 2
來源
第六屆河南省程序設計大賽
題目大意是：有n張牌，找到一個順序，使得第一次把上麵一張取出放到最下麵，
然後取出最上麵一張是一，第i次把上麵i張取出放到最下麵，然後取出一張是i；
//模擬即可
#include<stdio.h>
#include<string.h>
int a[20],b[20];
void MoveBack(int n,int m)//將第一張牌移到牌堆最後
{
int i;
for(i=1;i<m;i++)
{
a[i]=a[i+1];
}
a[m]=n;
}
void DelTop(int m)//刪除第一張牌
{
int i;
for(i=1;i<=m;i++)
{
a[i]=a[i+1];
}
}
int main()
{
int i,j,n,m,k,v,x,flag;
scanf("%d",&n);
while(n--)
{
scanf("%d",&m);
for(i=1;i<=m;i++)
a[i]=i;
x=1;k=1;flag=m;
while(m)
{
v=x;
while(v--)
MoveBack(a[1],m);
b[a[1]]=k++;
DelTop(m--);
x++;
}
printf("%d",b[1]);
for(i=2;i<=flag;i++)
printf(" %d",b[i]);
puts("");
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
}
return 0;
}
最後更新：2017-04-03 12:56:36