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Find if given number is sum of first n natural numbers in C++
In this problem, we are given a number num. Our task is to find if the given number is the sum of first n natural numbers.
Problem Description: Here, we need to check whether the given number is the sum of first n natural numbers.
Let’s take an example to understand the problem,
Input: num = 55
Output: yes, 10
Explanation:
55 is the sum of the first 10 natural numbers, 1+2+3+4+5+6+7+8+9+10.
Solution Approach:
A simple approach to solving the problem is finding the sum of n natural numbers until it becomes equal to or greater than num.
If the sum is equal to num, return n.
If at any value of n the sum becomes greater than n, return -1.
Program to illustrate the working of our solution,
Example
#include <iostream>
using namespace std;
int isNatSum(int num){
int sum = 0;
for (int n = 1; sum < num; n++) {
sum += n;
if (sum == num)
return n;
}
return -1;
}
int main(){
int num = 55;
int n = isNatSum(num);
if(n == -1)
cout<<"The value is not sum of natural numbers";
else
cout<<"The value is a sum of first "<<n<<" natural numbers";
return 0;
}Output −
The value is a sum of first 10 natural numbers
This method is good but we can solve the problem more efficiently using the mathematical formula for the sum of n natural numbers.
Sum of first mutual numbers is given by the formula,
sum = n*(n+1)/ 2
We are given sum and we need to find the value of n,
So we need to create a quadratic equation to find n.
=> 2*Sum = n2 + n
=> n2 + n - 2*sum = 0 , quadratic equation
The solution of this quadratic equation is,
Program to illustrate the working of our solution,
Example
#include <iostream>
#include <math.h>
using namespace std;
int isNatSum(int num){
int n = ( -1+ sqrt (1 + (8*num) ))/2;
if(ceil(n)==floor(n)){
return n;
}
return -1;
}
int main(){
int num = 55;
int n = isNatSum(num);
if(n == -1)
cout<<"The value is not sum of natural numbers";
else
cout<<"The value is a sum of first "<<n<<" natural numbers";
return 0;
}Output
The value is a sum of first 10 natural numbers
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