ConsecutivePrimeSum.cpp

// Consecutive Prime Sum

// Some prime numbers can be expressed as Sum of other consecutive prime numbers.

// For example

// 5 = 2 + 3 
// 17 = 2 + 3 + 5 + 7 
// 41 = 2 + 3 + 5 + 7 + 11 + 13

// Your task is to find out how many prime numbers which satisfy this property are present in the range 3 to N subject to a constraint that summation should always start with number 2.

// Write code to find out number of prime numbers that satisfy the above mentioned property in a given range.

// Input Format:
// Each test case contains a number N <= 1000000000

// Output Format:
// Print the total number of all such prime numbers which are less than or equal to N.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int isPrime(int n)
{
    for(int i=2;i<=(int)sqrt(n);i++)
    {
        if(n%i==0)
            return 0;
    }
    return 1;
}

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int i,j,n,ct=0,sum=2;
    cin>>n;
    for(i=3;sum<n;i+=2)
    {
        if(isPrime(i))
        {    
            sum+=i;
            if(isPrime(sum) && sum<=n)
                ct++;
        }
    }
    cout<<ct;
    return 0;
}