Problem:
Score : 100 points

Problem Statement

How many ways are there to choose two distinct positive integers totaling N, disregarding the order?

Constraints

1 \leq N \leq 10^6

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

4

Sample Output 1

1

There is only one way to choose two distinct integers totaling 4: to choose 1 and 3. (Choosing 3 and 1 is not considered different from this.)

Sample Input 2

999999

Sample Output 2

499999

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
 
int main(void){
  int n,i,t;
  scanf ("%d",&n);
  int d[n],a[n],b;
  for (i=0;i<n;i++) scanf ("%d",d[i]);
  for (i=0;i<n;i++) { /*全ての値に対して*/
    for (t=0;t<n-1;t++) if (d[i]==t+1) a[t]++;/*距離1,2,...の判定＆個数*/
                    }
  for (t=0;t<n-1;t++) {
  b*=a[t]/a[t+1];
  }
  printf("%d",b/2);
  return 0;
}