Codeforces Round #411 (Div. 2) C. Find Amir

C. Find Amir

time limit per test  

1 second

memory limit per test  

256 megabytes

 

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  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.

 

input

2

output

0

input

10

output

4

Note：

In the first example we can buy a ticket between the schools that costs .

 

题目大意：
　　　　　有n所学校，[1...n],从i学校到j学校需要花费（i+j）%（n+1）元，Sajjad想知道他参观完所有学校所需要的最少花费
解题思路：
　　　    本来想建个图用最小生成树写勒，但是发现可以利用（i+j）%（n+1）造出几条花费为0，并且两条花费为0的道路、

　　　　  连接的最小花费为1，所以只用判断一共可以造出几条花费为0的路即可。

　　　　　拿第二组样例来说，（1,10),(2，9),(3,8),(4,7),(5,6)这5条路的花费为0，用（10,2）来连接（1,10)和(2,9)这两条路花费为1...

　　　　　所以最后的花费为4.

　　　　　当n为奇数时情况有些不同（试着推一推）把奇数+1处理即可。

 

#include <stdio.h>
int main ()
{
    int n;
    while(~scanf("%d",&n))
    {
        printf("%d\n",(n+1)/2-1);
    } 
    return 0;
}