a, b, c = map(int, input().split(" "))
a = int(a)
b = int(b)
c = int(c)
lst = [a, b, c]
lst.sort()
d = lst[0]-lst[1]*lst[2]
e = lst[0]-lst[1]-lst[2]
if d<e:
    min = d
else:
    min = e
print(min)

#include <stdio.h>
int main(void) {
    int a, b, c, max, min, med;
    scanf("%d %d %d", &a, &b, &c);
    max = a >= b && a >= c? a: b >= a && b >= c? b: c;
    min = a <= b && a <= c? a: b <= a && b <= c? b: c;
    med = a + b + c - min - max;
    if (med == 0)
        printf("%d", - max);
    else
        printf("%d", min - max * med);
    return 0;
}

pleace

Answer is a - b - c or a - b * c

//We need to sort out 3 in the second power of the options for the arrangement of signs

#include <iostream>

using namespace std;

int main()
{
    int a, b, c, min;
    cin >> a >> b >> c;
    min = a + b + c;
    if(a + b - c < min) min = a + b - c;
    if(a - b + c < min) min = a - b + c;
    if(a - b - c < min) min = a - b - c;
    if(a - b + c < min) min = a - b + c;
    if(a - b * c < min) min = a - b * c;
    if(a * b * c < min) min = a * b * c;
    if(a * b + c < min) min = a * b + c;
    if(a * b - c < min) min = a * b - c;
    cout << min;
    return 0;
}

import java.util.Scanner;

public class Test{
    public static void main(String[] args) {

   Scanner sc = new Scanner(System.in);
    int a = sc.nextInt();
   int b = sc.nextInt();
   int c = sc.nextInt();
   if(a>=0 && b>=0 && c>=0 && a<=100 && b<=100 && c<=100 && a<=b && b<=c ){
       if(a==0 && b==0 || a==0 && b==1|| b==0||c==0||(a==1&& b==1&&c==1)){
    System.out.println(a-b-c);
    }
      else  System.out.println(a-b*c);
}
    }
}

#include <stdio.h>
int main(){
    int a,b,c;
    scanf("%d",&a);
    scanf("%d",&b);
    scanf("%d",&c);
    if (a>b){
        if (b>c){
            printf("%d",c-(a*b));
        }
        else{
            printf("%d",b-(a*c));
        }
    }
    if(b>a){
        if (a<c){
            printf("%d",a-(c*b));
        }
    }
return 0;
}

"The numbers are arranged in non-decreasing order (0 ≤ a ≤ b ≤ c ≤ 100)."
I my opinion, reading this we assume that the input will already be sorted.

The minimal output is: 1-2*5 = -9.

Edited by author 04.03.2020 22:45

import java.util.Arrays;
import java.util.Scanner;

public class ppppp {
    public static void main(String[] args) {

            Scanner in = new Scanner(System.in);
            int n = in.nextInt();
            int k = in.nextInt();
            int g = in.nextInt();

            if (0 <= n && n  <= k && k <= g && g <= 100){
            int[] mas = {n, k, g};
            boolean isSorted = false;
            int buf;
            while (!isSorted) {
                {
                    isSorted = true;
                    for (int i = 0; i < mas.length - 1; i++) {
                        if (mas[i] > mas[i + 1]) {
                            isSorted = false;

                            buf = mas[i];
                            mas[i] = mas[i + 1];
                            mas[i + 1] = buf;
                        }
                    }
                }

            }


            int da = mas[0] - mas[1] * mas[2];
            System.out.println(da);
        }
    }
}

lol

Edited by author 15.12.2019 12:57

Edited by author 15.12.2019 12:57

Edited by author 15.12.2019 12:57

var
a,b,c,s: longint;
begin
 readln(a,b,c);
 if (a > b) or (a > c) then
   if (b > c) then s:= c - a * b
    else
      s:= b - a * c;
 if (b > a) or (b > c) then
   if a > c then s:= c - a * b
    else
    s:= a - b * c;
  if (c > a) or (c > b) then
   if a > b then s:= b - a * c
    else
     s:=a - b * c;
 writeln(s);
end.

[code deleted]

This code Accepted so A can be 0 or 1; B can be 0 or 1; C can be 0 or 1 in tests.
But if we will see on combinatons then there are only 4 combinations (if a<=b<=c) and there are 8 tests as min.

The paradox

Edited by author 14.03.2016 01:18

Edited by moderator 24.11.2019 13:45

little hint: if n=m=0 or n=m=1 or n=0 and m=1 you should use n-m-l expression ;)

Can I get some test inputs please?