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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include <string.h>
#include<algorithm>
#include<math.h>
using namespace std;
int main(void){
long int N;
int i,cnt,count;
cnt=0;
scanf("%ld",&N);
long int A[N], A_origin[N],B_origin[N], B[N];
for (i=0;i<N;i++){
	scanf("%ld", &A_origin[i]);
	A[i]=A_origin[i];
}
for (i=0;i<N;i++){
	scanf("%ld", &B_origin[i]);
	B[i]=B_origin[i];
}

sort(A, A+N);
sort(B, B+N);
for (i=0;i<N;i++){
if (A[i]>B[i]){
	cnt++;	
}
if (A_origin[i]>B_origin[i]){
	count++;	
}


}
if(count>N-2 || cnt>0){
	printf("NO");
}
else {
	printf("YES");
}
}