- Related Questions & Answers
- Count ways to express ‘n’ as sum of odd integers in C++
- Sum of the Series 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ... in C++\n
- Sum of the series 1 + (1+2) + (1+2+3) + (1+2+3+4) + ... + (1+2+3+4+...+n) in C++
- Count ways to express a number as sum of powers in C++
- Program to find sum of series 1*2*3 + 2*3*4+ 3*4*5 + . . . + n*(n+1)*(n+2) in C++
- C++ program to find the sum of the series 1/1! + 2/2! + 3/3! + 4/4! +…….+ n/n!
- Count ways to express a number as sum of consecutive numbers in C++
- C++ Program to find the sum of a Series 1/1! + 2/2! + 3/3! + 4/4! + …… n/n!
- C++ program to find the sum of the series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + … + (n*n)
- Program to find sum of series 1 + 1/2 + 1/3 + 1/4 + .. + 1/n in C++
- Python Program to find the sum of a Series 1/1! + 2/2! + 3/3! + 4/4! +…….+ n/n!
- Java Program to find the sum of a Series 1/1! + 2/2! + 3/3! + 4/4! +…….+ n/n!
- Count number of trailing zeros in (1^1)*(2^2)*(3^3)*(4^4)*.. in C++
- C++ program to find the sum of the series 1 + 1/2^2 + 1/3^3 + …..+ 1/n^n
- Sum of the series 1^1 + 2^2 + 3^3 + ... + n^n using recursion in C++

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

Given a positive number N as input. The goal is to find the number of ways in which we can express N as a sum of 1s, 3s and 4s only. For example, if N is 4 then it can be represented as 1+1+1+1, 3+1, 1+3, 4 so the number of ways will be 4.

**For Example**

**Input -** N=5

**Output -** Count of different ways to express N as the sum of 1, 3 and 4 are: 6

**Explanation -** 5 can be represented as:

- 1+1+1+1+1
- 1+3+1
- 3+1+1
- 1+1+3
- 4+1
- 1+4

**Input -** N=6

**Output -** Count of different ways to express N as the sum of 1, 3 and 4 are: 9

**Explanation -** 9 can be represented as:

- 1+1+1+1+1+1
- 3+1+1+1
- 1+3+1+1
- 1+1+3+1
- 1+1+1+3
- 3+3
- 4+1+1
- 1+4+1
- 1+1+4

In this approach we will use a dynamic programming approach to count the number of ways we can represent N as sum of 1, 3 and 4. Take array arr[i] where i represents the number and arr[i] as ways to represent it as sum.

The base cases will be

arr[0]=0 ( No way )

arr[1]=1 ( only one way , 1 )

arr[2]=1 ( only one way, 1+1 )

arr[3]=2 ( 1+1+1, 3 )

Now other numbers 4, 5 … i etc. will have ways as arr[i-1]+arr[i-3]+arr[i-4].

- Take a positive number N as input.
- Function Expres_sum(int N) takes N and returns the count of different ways to express N as the sum of 1, 3 and 4.
- Take array arr[N+1] to store the count of ways.
- Initialize base cases arr[0] = 1, arr[1] = 1, arr[2] = 1 and arr[3] = 2.
- Traverse for rest of the values from i=4 to i<=N.
- Takeaways arr[i] as sum of arr[i - 1] + arr[i - 3] + arr[i - 4].
- At the end of the for loop return arr[N] as result.

#include <bits/stdc++.h> using namespace std; int Expres_sum(int N) { int arr[N + 1]; arr[0] = 1; arr[1] = 1; arr[2] = 1; arr[3] = 2; for (int i = 4; i <= N; i++) { arr[i] = arr[i - 1] + arr[i - 3] + arr[i - 4]; } return arr[N]; } int main() { int N = 5; cout << "Count of different ways to express N as the sum of 1, 3 and 4 are: " << Expres_sum(N); return 0; }

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

Count of different ways to express N as the sum of 1, 3 and 4 are: 6

Advertisements