How one can mathematically prove that it is always optimal to send people in one direction with max number of seats?

Why I don't get AC
here is my code:

var n,k,m,AC:integer;
begin
readln(n,k);
m:=n div 2;
if k=m then AC:=0;
if k<m then AC:=m-k;
if k>m then AC:=k-m-1;
write(AC);
end.

#include<stdio.h>


int n,k,answer;
int main()
{
scanf("%d %d",&n,&k);
int len = n/2;


if(k==len)
{
printf("0");
return 0;
}
else if(k<len)
{
answer = len - k;
}
else
{
answer = k - len - 1;
}

printf("%d",answer);
}

Very easy:)

*first if k>n/2 make it smaller n/2: k=n-k+1

*then the answer is (n-k-2)

 (......[Vasya].......[n-1][n]
  if [n-1] and [n] sit,then all people between
  [Vasya] and [n-1] will stumble at Vasya.
  And this will be the maximal answer )

*special case: n=2 => answer is 0

i had that problem and solved it.
2 nd test is n=2 and answer is 0;

//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("avx")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

#define re return
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define sqrt(x) sqrt(abs(x))
#define mp make_pair
#define pi (3.14159265358979323846264338327950288419716939937510)
#define fo(i, n) for(int i = 0; i < n; ++i)
#define ro(i, n) for(int i = n - 1; i >= 0; --i)
#define unique(v) v.resize(unique(all(v)) - v.begin())

template <class T> T abs (T x) { re x > 0 ? x : -x; }
template <class T> T sqr (T x) { re x * x; }
template <class T> T gcd (T a, T b) { re a ? gcd (b % a, a) : b; }
template <class T> int sgn (T x) { re x > 0 ? 1 : (x < 0 ? -1 : 0); }

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef double D;
typedef long double ld;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef unsigned long long ull;
typedef tree <pair<int, char>, null_type, less<pair<int, char>>, rb_tree_tag, tree_order_statistics_node_update> _tree;

int main() {
    int n, k;
    cin >> n >> k;
    cout << max(0, max(n - k - 2, k - 3)) << endl;
}

import java.util.Scanner;

public class Sold_Out {
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);

        int n = scan.nextInt();
        int k = scan.nextInt();

        if(n<3){
            System.out.println(0);
            return;
        }

        System.out.println(Math.max(k - 1, n - k) - 2);
    }
}

#include "stdio.h"
int main()
{
    int n,i;
    scanf("%d %d",&n,&i);
    if(n==2)
        {printf("0");
    return 0;
    }
    if(i<=n/2)
    printf("%d",n-i-2);
    else
        printf("%d",i-1-2);

}

Please help me .this code failed at test case 5
#include <stdio.h>
int main() {
int n,k;
scanf ("%d\n%d" ,&n,&k);
int l, pos;
 l = n % 2 ;

if ( l != 0 || n < 1 || k < 1 || k > n     || n > 50 )
{
   printf("invalid input");
}
else
{
   if (n != 2 )
   {
 if ( k <= n/2  )
   {
   pos  = n/2 - k ;
   printf("%d", pos);
   }
   else
   {
   pos = k - ((n/2) + 1 );
    printf("%d", pos);
   }
    }
}
if ( n == 2 )
{
  pos = 0;
  printf("%d", pos);

}
return 0;
}

# include <iostream>
using namespace std;
int main ()
{
int n,k,sol;
    cin>>n>>k;
        if(int(n/2) && 1<=k<=n<=50){
        if(k<=n/2) sol=n-3;
        if(k>n/2) sol=k-3;
        if(n==2) sol=0;
        }
    cout<<sol<<endl;
}

Edited by author 26.06.2014 22:00

var a,b,n,k:integer;
begin
readln(n,k);
a:=n div 2;
if k>a then b:=n-k
       else if (k=a) or (k=a+1) then b:=0
                                else b:=a-k;
writeln(b)
end.

where is my mistake? falls on 5th

var i,nr,n,k,c:integer;
begin
read(n,k);
nr:=trunc(n/2);
if (k>nr) then
begin
for i:=nr+1 to k-1 do
inc(c);
end;
if (k<nr) then
begin
for i:=k+1 to nr do
inc(c);
end;
write(c);
end.

where is my mistake? falls on 5th

var i,nr,n,k,c:integer;
begin
read(n,k);
nr:=trunc(n/2);
if (k>nr) then
begin
for i:=nr+1 to k-1 do
inc(c);
end;
if (k<nr) then
begin
for i:=k+1 to nr do
inc(c);
end;
write(c);
end.

how name type this problem, type, where you should create 1-2 formuli and choise minimal or maximal value in one of it?

Suppose
* n in uneven,
* the first (n-1)/2 seats and the last (n-1)/2 seat has the same number of people already sitting and
* the person sitting in seat (n+1)/2 arrives.
The description does not say if he will choose left or right.
I got AC by assuming he will not choose the same side as Vasya.

No subject

Edited by author 18.07.2011 13:04

Edited by author 18.07.2011 13:04

I don't know where I made mistake,it falls on 5th example!!!


# include <iostream>
using namespace std;
int n,k,m,sol;
int main ()
{
    cin>>n>>k;
    m=n/2;
    if (k==m) sol=0;
    if (k<m) sol=m-k;
    if (k>m) sol=k-m-1;
    cout<<sol<<endl;
    system("pause");
}