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# Count numbers which can be constructed using two numbers in C++

We are provided three numbers X, Y and N ( to define range [1,N] ). The goal is to find all the numbers in the range [1,N] that can be constructed using X and Y only any number of times..

For example if X=2 and Y=3. Number 6 can be constructed using 2 thrice ( 2+2+2 ) or 3 twice ( 3+3 ). Similarly 7 can be constructed using 2 twice and 3 once ( 2+2+3 ).

We will do this by subtracting X or Y from each number from 1 to N. If the final number reduces to 0 then increment count.

Let’s understand with examples.

**Input**

N=10 X=4, Y=3

**Output**

Total numbers constructed using X & Y only: 7

**Explanation**

Numbers constructed by 3 and 4 only: 3, 4, 6 (3+3), 7 (3+4), 8 (4+4), 9 (3+3+3), 10 (3+3+4)

**Input**

N=10 X=5, Y=4

**Output**

Total numbers constructed using X & Y only: 5

**Explanation**

Numbers constructed by 4 and 5 only: 4, 5, 8(4+4), 9 (4+5), 10 (5+5)

## Approach used in the below program is as follows

We take three integers X, Y and N.

Function constructNums(int n,int x,int y) returns the count of numbers which can be constructed using x and y only.

Take the initial variable count as 0 for such numbers.

Traverse range of numbers using for loop. i=1 to i<=n

Now for each number num=i, using while loop check if num>0,

while(num%x==0) and num is not 0 subtract x from it.

while(num%y==0) and num is not 0 subtract y from it.

If after subtracting x and y number is non divisible by both and is greater than both.Subtract both x and y from it.

If after the outer while loop check if num is 0. Means x, y or both can make num. Increment count.

At the end of all loops count will have a total number.

Return the count as result.

## Example

#include <bits/stdc++.h> using namespace std; int constructNums(int n,int x,int y){ int count = 0; for (int i = 1; i <= n; i++) { int num = i; while(num>0){ while((num%x)==0 && num!=0) num-=x; while((num%y)==0 && num!=0) num-=y; if (num>x && num>y) num=num-x-y; else break; } if (num==0) { count++;} } return count; } int main(){ int N=20; int X=5,Y=4; cout <<"Total numbers constructed using X & Y only:"<<constructNums(N,X,Y); return 0; }

## Output

If we run the above code it will generate the following output −

Total numbers constructed using X & Y only:14

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