Write a program that takes an integer and prints the number of trailing zeroes. Example: Enter the number: 24100
Trailing zeroes: 2

I have no Idea what condition to create to determine the number of zeroes in a number.

CodePudding user response：

You could use a modulus trick here:

int input = 24100;
int num_zeroes = 0;

while (input > 0) {
    if (input % 10 == 0) {
        num_zeroes = num_zeroes   1;
        input = input / 10;
    }
    else {
        break;
    }
}

printf("%d", num_zeroes);  // 2

CodePudding user response：

actually you didn't mention the definition of a trailing zero , but from the definition I found online like here.

In mathematics, trailing zeros are a sequence of 0 in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

meaning that 12500 has 2 trailing zeros , also 125.50000 has 4 trailing zeros and so on , which means that trailing zero is just number of zeros at the end of the decimals after which no other digit follows , so the easiest way to do is take string from user and compare every character with its ascii representation to find number of consecutive zeros and here the code:

    #include <stdio.h>

    int main(void)
    {
        char input_string[100];
        printf("enter a number : ");
        scanf("%s", &input_string);
        int numberOfTrailingZeros = 0;

        for (int i = 0; input_string[i] != '\0' ;   i) {
            if(input_string[i] == '0')
                numberOfTrailingZeros  ;
            else if(input_string[i] == '.')
                continue;
            else
                numberOfTrailingZeros = 0;
        }
        printf("number of trailing zeros = %d\n", numberOfTrailingZeros);

    }