CF411div2C.Find Amir

最新推荐文章于 2017-10-23 16:08:15 发布

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本文探讨了在一个由多个学校构成的网络中寻找一条访问所有学校的最经济路径的问题。通过一种特殊的访问顺序，即从第一个学校开始，交替访问两端的学校，可以有效地减少所需购买票券的数量，从而实现最低成本遍历。

A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends.

There are n schools numerated from 1 to n. One can travel between each pair of them, to do so, he needs to buy a ticket. The ticker between schools i and j costs $\text{[math]}$ and can be used multiple times. Help Sajjad to find the minimum cost he needs to pay for tickets to visit all schools. He can start and finish in any school.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of schools.

Output

Print single integer: the minimum cost of tickets needed to visit all schools.

题目大意： 1-n个城市每个城市走一边找出一条路线使得路线 $\text{[math]}$ 之和最小

方案从1走到n n走到2 2走到n-1 n-1走到3 3走到n-2 依次这样走没走两步结果加一

 #include<iostream>
using namespace std;
int n;
int main()
{
	 cin>>n;
	 cout<<(n+1)/2-1;
	 return 0;
 }