CF#804 A. Find Amir（贪心）

C. Find Amir time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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.

Examples input 2 output 0 input 10 output 4 Note

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

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); } }