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Count ways to express ‘n’ as sum of odd integers in C++
Given an integer n as input. The goal is to find the number of ways in which we can represent ‘n’ as the sum of odd integers. For example, if n is 3 it can be represented as sum ( 1+1+1 ) and (3) so total 2 ways.
For Example
Input
n=6
Output
Count of ways to express ‘n’ as sum of odd integers are: 8
Explanation
The ways in which we can express ‘n’ as sum of odd integers − 1. 1+1+1+1+1+1 2. 3+1+1+1 3. 1+3+1+1 4. 1+1+3+1 5. 1+1+1+3 6. 3+3 7. 1+5 8. 5+1
Input
n=9
Output
Count of ways to express ‘n’ as sum of odd integers are: 34
Explanation
The some of the ways in which we can express ‘n’ as sum of odd integers: 1. 1+1+1+1+1+1+1+1+1 2. 3+3+3 3. 5+3+1 4. 7+1+1 5. ….and other such combinations
Approach used in the below program is as follows −
In this approach we will check the ways of representing a number as a sum of odd integers from previous numbers that are n−1th and n−2th numbers. Ways will be ways(n−1) + ways(n−2).
Take an integer n as input.
Function odd_ways(int n) takes a number and returns the count of ways to express ‘n’ as sum of odd integers.
Take an array arr of length n+1 to store count ways for representing numbers as sum of odd integers.
For number 0, there is no such way so set arr[0] with 0.
For number 1 there is only one way so set arr[1] with 1.
For the rest of numbers we can set arr[i] with arr[i−1]+arr[i−2] for i between 2 to n.
At the end we have arr[n] for the number of ways in which n is represented as the sum of odd integers.
Return arr[n] as result.
Example
#include<iostream>
using namespace std;
int odd_ways(int n){
int arr[n+1];
arr[0] = 0;
arr[1] = 1;
for(int i = 2; i <= n; i++){
arr[i] = arr[i-1] + arr[i-2];
}
return arr[n];
}
int main(){
int n = 6;
cout<<"Count of ways to express ‘n’ as sum of odd integers are: "<<odd_ways(n);
return 0;
}Output
If we run the above code it will generate the following output −
Count of ways to express ‘n’ as sum of odd integers are: 8
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