ZOJ 1730 Crazy Tea Party
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n participants of "crazy tea party" sit around the table. Each minute one pair of neighbors can change their places. Find the minimum time (in minutes) required for all participants to sit in reverse order (so that left neighbors would become right, and right - left).
Input. The first line is the amount of tests. Each next line contains one integer n (1 <= n <= 32767) - the amount of crazy tea participants.
Output. For each number n of participants to crazy tea party print on the standard output, on a separate line, the minimum time required for all participants to sit in reverse order.
Sample Input
3
4
5
6
Sample Output
2
4
6
题意:几个人围成圈,每分钟相邻的2个人可以对调,问最少要用多少时间使每个人原先左右两边的人对调
思路:先只考虑排成一列的情况。由于只能在邻居之间交换。相当于用冒泡排序排列n..1,需(n-1)n/2(∵T(n) = n-1 + T(n))
然后考虑将环转换为非环。即(k k-1 .. 1) (n .. k+1)例如:1 2 3 4 5 6的目标可以是 6 5 4 3 2 1,也可以是 (4 3 2 1 )(6 5)。
相当于进行2个冒泡排序,2个冒泡排序的n相等时,和最小
#include <stdio.h>
int main()
{
int M;
scanf("%d",&M);
while(M--){
int n,k;
scanf("%d",&n);
if(n%2==0)
k=(n/2-1)*(n/2);
else
k=((n-1)/2)*((n-1)/2-1)/2+((n+1)/2)*((n+1)/2-1)/2;
printf("%d\n",k);
}
return 0;
}
int main()
{
int M;
scanf("%d",&M);
while(M--){
int n,k;
scanf("%d",&n);
if(n%2==0)
k=(n/2-1)*(n/2);
else
k=((n-1)/2)*((n-1)/2-1)/2+((n+1)/2)*((n+1)/2-1)/2;
printf("%d\n",k);
}
return 0;
}
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