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Count of numbers from range[L, R] whose sum of digits is Y 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 sum of digits equal to a given number Y.
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. If this sum is equal to Y, increment count.
Let’s understand with examples.
Input
START=10 END=20 Y=4
Output
Numbers such that digit sum is equal to Y: 1
Explanation
Number 13 has digit sum equal to 4.
Input
START=10 END=50 Y=5
Output
Numbers such that digit sum is equal to Y: 5
Explanation
Numbers 14, 23, 32, 41 and 50 have digit sum 5.
Approach used in the below program is as follows
We take integers START, END, Y.
Function digitSum(int start, int end, int y) returns the count of numbers with digitsum=y
Take the initial variable count as 0 for such numbers.
Take variable digsum as 0
Traverse range of numbers using for loop. i=start to i=end
Now for each number num=i, using while loop check if number is >0.
calculate digsum+=num%10. Reduce num=num/10 to add the next digit.
At the end of the while, check if ( digsum == d ). If true increment count.
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 y){
int count = 0;
int digsum = 0;
for (int i = start; i <= end; i++){
int num=i;
digsum=0;
while(num>0){
digsum+=num%10; //sum of digits
num=num/10;
}
if(digsum==y) //original number is i{
count++;
//cout<<i<<" ";
}
}
return count;
}
int main(){
int START = 100;
int END = 1000;
int Y = 5;
cout <<"Numbers such that digit sum is equal to Y: "<<digitSum(START,END,Y);
return 0;
}Output
If we run the above code it will generate the following output −
Numbers such that digit sum is equal to Y: 15
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