Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M in C++
We are provided two numbers START and END to define a range of numbers.The goal is to find all the numbers in the range [START,END] which have no digit as 0 and have sum of digits equal to a given number N. Also the numbers are divisible by M
We will do this by traversing numbers from START to END and for each number we will count the sum of its digit using a while loop ( only if all digits are non zero ). If this sum is equal to N and the number is divisible by M, increment count.
Let’s understand with examples.
Input
START=1 END=100 N=9 M=6
Output
Numbers with digit sum N and divisible by M: 4
Explanation
Numbers 18, 36, 54, 72 have digit sum=9 and divisible by 6. None has 0 as a digit.
Input
START=100 END=200 N=10 M=2
Output
Numbers with digit sum N and divisible by M: 4
Explanation
Numbers 118, 136, 154, 172 have digit sum=10 and divisible by 2. None has 0 as a digit.
Approach used in the below program is as follows
We take integers START, END, N and M.
Function digitSum(int start, int end, int n, int m) returns the count of numbers with digitsum=n and divisible by m and have all non-zero digits.
Take the initial variable count as 0 for such numbers.
Take variable digsum as 0
Take variable flag as 0.
Traverse range of numbers using for loop. i=start to i=end
Now for each number num=i, if num%m==0 ( divisible by m ) move forward.
using while loop check if number is >0. And find digits.
digit=num%10. If digit is non-zero calculate digsum+=digit. Reduce num=num/10 to add the next digit. If any digit is 0 set flag=0 and break the while loop
At the end of the while, check if ( digsum == n and flag==1 ). If true increment count.
Now increment i by m ( in multiples of m ).
At the end of all loops count will have a total number which satisfies the condition.
Return the count as result.
Example
#include <bits/stdc++.h>
using namespace std;
int digitSum(int start, int end, int n, int m){
int count = 0;
int digsum = 0;
int flag=0;
for (int i = start; i <= end; i++){
int num=i;
digsum=0;
flag=0;
if(num%m==0){
while(num>0){
int digit=num%10;
if(digit==0){
flag=0;
break;
}
digsum+=num%10; //sum of digits
num=num/10;
flag=1;
}
if(digsum==n && flag==1) //original number is i {
count++;
cout<<i<<" ";
}
i+=m; //now increment in multiples of m
i--; // for loop has i++
}
}
return count;
}
int main(){
int START = 1;
int END = 100;
int N = 9;
int M = 6;
cout <<"Numbers with digit sum N and divisible by M: "<<digitSum(START,END,N, M);
return 0;
}Output
If we run the above code it will generate the following output −
Numbers with digit sum N and divisible by M: 4
282 Views
