Problem:
Score : 100 points

Problem Statement

Snuke lives at position x on a number line.
On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery.

Snuke decided to get food delivery from the closer of stores A and B.
Find out which store is closer to Snuke's residence.

Here, the distance between two points s and t on a number line is represented by |s-t|.

Constraints

1 \leq x \leq 1000

1 \leq a \leq 1000

1 \leq b \leq 1000

x, a and b are pairwise distinct.

The distances between Snuke's residence and stores A and B are different.

Input

Input is given from Standard Input in the following format:

x a b

Output

If store A is closer, print A; if store B is closer, print B.

Sample Input 1

5 2 7

Sample Output 1

B

The distances between Snuke's residence and stores A and B are 3 and 2, respectively.
Since store B is closer, print B.

Sample Input 2

1 999 1000

Sample Output 2

A

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

typedef long long ll;
typedef int keytype;

int gcd(int a, int b){ return (a % b) ? gcd(b, a % b) : b;}
int lcm(int a, int b){ return (a / gcd(a, b)) * b;}

void quicksort(keytype a[], int first, int last)
{
    int i, j;
    keytype x, t;

    x = a[(first + last) / 2];
    i = first;  j = last;
    for ( ; ; ) {
        while (a[i] < x) i++;
        while (x < a[j]) j--;
        if (i >= j) break;
        t = a[i];  a[i] = a[j];  a[j] = t;
        i++;  j--;
    }
    if (first  < i - 1) quicksort(a, first , i - 1);
    if (j + 1 < last) quicksort(a, j + 1, last);
}

int is_prime(int x){
    if(x <= 1) return 0;
    for(int i = 2; i * i <= x; i++){
        if(x % i == 0) return 0;
    }
    return 1;
}

int is_substr(char s[], char t[], int i){
    int flag = 1;

    if(s[i] == '\0') flag = 0;

    for(int j = 0; j < strlen(t); ++j, ++i){
        if(s[i] != t[j]){
            flag = 0;
            break;
        }
    }
    return flag;
}

// factor list, number of factors, natural number
void factorize(int ftr[], int *num, int n){
    int a = 2, cnt = 0;

    while(n >= a){
        if(n % a == 0){
            n = n / a;
            ftr[cnt++] = a;
        }else{
            a++;
        }
    }
    *num = cnt;
}

void comb(int size1, int size2, int v[size1][size2]){
  for(int i = 0; i < size1; i++){
    v[i][0] = 1;
    v[i][i] = 1;
  }
  for(int k = 1; k < size1; k++){
    for(int j = 1; j < k; j++){
      v[k][j] = (v[k-1][j-1] + v[k-1][j]);
    }
  }
}

int abs_val(int a, int b){
    if(a > b) return a - b;
    else return b - a;
}

int main(){
    int x, a, b;

    scanf("%d %d %d", &x, &a, &b);

    int dis_a = abs_val(x, a);
    int dis_b = abs_val(x, b);

    if(dis_a < dis_b) puts("A");
    else puts("B");

    return 0;
}