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Count numbers having 0 as a digit in C++
We are provided a number N. The goal is to find the numbers that have 0 as digit and are in the range [1,N].
We will do this by traversing numbers from 10 to N ( no need to check from 1 to 9 ) and for each number we will check each digit using a while loop. If any digit is found as zero increment count and move to next number otherwise reduce the number by 10 to check digits until number is >0.
Let’s understand with examples.
Input
N=11
Output
Numbers from 1 to N with 0 as digit: 1
Explanation
Starting from i=10 to i<=11 Only 10 has 0 as a digit. No need to check the range [1,9].
Input
N=100
Output
Numbers from 1 to N with 0 as digit: 10
Explanation
10, 20, 30, 40, 50, 60, 70, 80, 90, 100. Ten numbers have 0 as digits.
Approach used in the below program is as follows
We take an integer N.
Function haveZero(int n) takes n as parameter and returns count of numbers with 0 as digit
Take the initial variable count as 0 for such numbers.
Traverse range of numbers using for loop. i=10 to i=n
Now for each number num=i, using while loop check if num%10==0, if false divide num by 10 and move to next digit until num>0
If true stop further checking, increment count and break while loop.
At the end of all loops count will have a total numbers with 0 as digit between 1 to N.
Return the count as result.
Example
#include <bits/stdc++.h>
using namespace std;
int haveZero(int n){
int count = 0;
for (int i = 1; i <= n; i++) {
int num = i;
while(num>1){
int digit=num%10;
if (digit == 0){
count++;
break;
}
else
{ num=num/10; }
}
}
return count;
}
int main(){
int N = 200;
cout <<"Numbers from 1 to N with 0 as digit: "<<haveZero(N);
return 0;
}Output
If we run the above code it will generate the following output −
Numbers from 1 to N with 0 as digit: 29
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