You can iterate through the digits as follows:
int digitToSearch = 3;
int count = 0;
while (number != 0)
{
int digit = number % 10;
if (digit == digitToSearch)
count++;
number /= 10;
}
문제
Ok, so at first, I'm very beginner in programming. It's my school homework and I cannot use conversion to string. Just if,else,for,while.
On input there is the number and the digit.
I know how to get information what number is a specified digit in a number but I have no idea how to find out how many of these numbers are there.
Let's say I have number 123 467 (it has to be less than 999 999) and I want the third number. I know it's bigger than 100 000, so I do the math - (int) 123 467 / 100 = 123 and then 123%10 = 3. Now I need to know if there are any more 3's in the number - but here is the point - I'm not sure what cycle should I use.
And I also have to create some code which determines how large is the number (bigger than 100/1000/10000/...).
I'm not asking for a full solution but little help would be appreciated. Even in a pseudolanguage.
Current code (almost nothing):
double digit, number;
try
{
digit = Convert.ToInt32(poledigit.Text);
number = Convert.ToInt32(polenumber.Text);
}
catch
{
MessageBox.Show("Zadejte číslo ve správném formátu");
return;
}
if (digit > 6 & number > 999999)
{
MessageBox.Show("Číslo musí být menší než 999 999 a digit musí být menší než 6.");
return;
}
while(number >= 100000)
{
number /= Math.Pow(10, digit);
number %= 10;
}
해결책 2
You can iterate through the digits as follows:
int digitToSearch = 3;
int count = 0;
while (number != 0)
{
int digit = number % 10;
if (digit == digitToSearch)
count++;
number /= 10;
}
다른 팁
I would create an int array counting the number of digits
int[] digitCount = new int[10]; // Range: digitCount[0..9]
Then determine the digits one by one by eliminating the last one, until the number is zero. The loop would repeat the following code:
int digit = number % 10;
number /= 10;
digitCount[digit]++;
Now digitCount
contains the count of each digit
int countOfDigit3 = digitCount[3];
If you cannot use arrays, count only the occurences of the desired digit
int digit = ...;
int digitCount = 0;
while (number != 0) {
int d = number % 10;
number /= 10;
if (d == digit) {
digitCount++;
}
}