fjnu 1655 Crazy tea party

最新推荐文章于 2024-08-20 10:12:01 发布

原创最新推荐文章于 2024-08-20 10:12:01 发布 · 824 阅读

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#output #iostream #each #integer #input #pair

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探讨了在疯狂茶会上，参与者围绕圆桌而坐，通过相邻两人互换位置的方式，所有人达到反向就座所需的最短时间。采用类似于冒泡排序的方法，将问题简化为环形数组的逆序问题。

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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

Sample Output

2
4
6

KEY:一看题目无处下手，认真分析可知，只要将环合理的拆开就好做了，比如123456 拆成123 456；要逆序只能交换，类似冒泡排序的方法；次数n*（n-1）/2；

  
Source:#include<iostream>
using namespace std;

int crazy(int n)
...{
    int t=n/2;
    if(n%2==0) return t*(t-1);
    else return (t+1)*t/2+t*(t-1)/2;
}

int main()
...{
    int N;
    cin>>N;
    int n;
    for(int i=1;i<=N;i++)
    ...{
        cin>>n;
        cout<<crazy(n)<<endl;
    }
    return 0;
}