POJ1455 Crazy tea patry

最新推荐文章于 2019-07-22 04:50:19 发布

crystal__love

最新推荐文章于 2019-07-22 04:50:19 发布

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分类专栏：算法文章标签： poj

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本文探讨了在一个环形序列中使所有参与者按逆序排列所需的最小时间问题。通过将环形序列转换为两个非环形序列进行分析，并提供了具体的算法实现。

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Description
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

本题意思是：环形序列逆序排列

首先想到冒泡的方法，如果不是环形，就能直接用冒泡的式子：即为n*（n-1）*0.5

现在的难点在于是环形，想到将环形转化为2个非环形

如果n是偶数，则两部分相等，均为0.5*n

如果n是奇数，则分成（n-1）*0.5和（n-1）*0.5+1

#include<stdio.h>
#include<stdlib.h>
int main()
{
int N,n;
int i;
scanf("%d",&N);
for(i=0;i<N;i++)
{
int ans;
scanf("%d",&n);
if(n%2==0)
ans=(n/2)*(n/2-1);
else
ans=0.5*(n/2)*(n/2-1)+0.5*(n/2+1)*(n/2);
printf("%d\n",ans);
}
return 0;
}