Difference#941

Difference (Editorial)

Algorithm Analysis

Let $n$ be the input number. Let $a$ be the number of hundreds, $b$ be the number of tens, $c$ be the number of ones. Then the answer will be the value of $a \times b \times c - (a + b + c)$ a \times b \times c - (a + b + c).

Algorithm Implementation

Read the value of $n$ , calculate its absolute value.

scanf("%d",&n);   
if (n < 0) n = -n;

Extract from the number $n$ the number of hundreds $a$ , tens $b$ , and ones $c$ .

a = n / 100;
b = (n / 10) % 10;
c = n % 10;

Calculate and output the answer.

res = a * b * c - (a + b + c);
printf("%d\n",res);

Implementation with Formatted Input

The first character can be a minus, read it. If there is no minus, then nothing will be read into ch.

#include <stdio.h>

int a, b, c, res;
char ch;

int main(void)
{
    scanf("%[^0-9]",&ch);
    scanf("%1d%1d%1d",&a,&b,&c);
    res = a * b * c - (a + b + c);
    printf("%d\n",res);   
    return 0;
}

Java Implementation

import java.util.*;

public class Main
{
    public static void main(String[] args)
    {
        Scanner con = new Scanner(System.in);
        int n = con.nextInt();
        if (n < 0) n = -n;
        int a = n / 100;
        int b = (n / 10) % 10;
        int c = n % 10;
        int res = a * b * c - (a + b + c);
        System.out.println(res);      
        con.close();
    }
}

Java

Python Implementation

n = int(input())
if n < 0: n = -n

a = n // 100
b = (n // 10) % 10
c = n % 10

res = a * b * c - (a + b + c)
print(res)

Python