CF#804 A. Find Amir（贪心）

最新推荐文章于 2022-07-05 02:07:52 发布

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本文探讨了一个有趣的问题：如何以最低的成本遍历n个学校。通过分析票价规律，即从i学校到j学校的票价为(i+j)%(n+1)，文章提出了一种策略，通过巧妙选择路径来最小化总费用。

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.

Examples

input
2

output
0

input
10

output
4

Note

In the first example we can buy a ticket between the schools that costs $\text{[math]}$ .

题意：有 1 ~ n 的 n 个车站，从 i 车站到 j 车站的票价为 ( i + j ) % ( n + 1 ) ，问我们至少要花费多少钱可以遍历所有车站。

思路：尽量使 i + j == n + 1即可，例 n = 10 ，我们可以选择（ 1 ， 10 ），（ 2， 9 ），（ 3 ， 8 ），（ 4 ， 7 ），（ 5 ， 6 ），（ 2 ， 10 ），（ 3 ， 9 ），（ 4 ， 8 ），（ 5 ， 7 ）易得规律 ans = n / 2 - （( n & 1 ) ? 0 : 1）。

#include <bits/stdc++.h>
using namespace std;

const int N = 2e5 + 10;
char a[N];
int n;

int main()
{
    while(scanf("%d", &n) == 1)
    {
       int ans = n / 2 - ((n & 1) ? 0 : 1);
       printf("%d\n", ans);
    }
}