# Print all n-digit numbers whose sum of digits equals to given sum in C++

In this problem, we are given two numbers n and sum. We have to print all n digit numbers whose sum is equal to the sum. In this problem, numbers with leading 0’s are not considered.

Let’s take an example to understand the problem,

Input: n = 2 , sum = 5
Output: 14 23 32 41 50
Explanation: The sum of digits of the number in all numbers in 5.

To solve this problem, we will have to find all the n-digit numbers with sum with the given sum value. For this, we will fix a digit place with all values and based on its position to be even or odd, call for values at other places in the number such that the condition remains satisfied.

## Example

Program to implement the above solution −

Live Demo

#include <iostream>
using namespace std;
void PrintNumberWithDigitSum(int n, int sum, char* out, int index) {
if (index > n || sum < 0)
return;
if (index == n) {
if(sum == 0) {
out[index] = ' ';
cout << out << " ";
}
return;
}
for (int i = 0; i <= 9; i++) {
out[index] = i + '0';
PrintNumberWithDigitSum(n, sum - i, out, index + 1);
}
}
void numberWithSum(int n, int sum) {
char out[n + 1];
for (int i = 1; i <= 9; i++) {
out = i + '0';
PrintNumberWithDigitSum(n, sum - i, out, 1);
}
}
int main() {
int n = 3, sum = 6;
cout<<"All "<<n<<" digit numbers with sum "<<sum<<" are :\n";
numberWithSum(n, sum);
return 0;
}

## Output

All 3 digit numbers with sum 6 are −
105 114 123 132 141 150 204 213 222 231 240 303 312 321 330 402 411 420 501 510 600