# Find the prime numbers which can written as sum of most consecutive primes

Given an array of limits. For every limit, find the prime number which can be written as the sum of the most consecutive primes smaller than or equal to limit.

The maximum possible value of a limit is 10^4.

Example:

```Input  : arr[] = {10, 30}
Output : 5, 17
Explanation : There are two limit values 10 and 30.
Below limit 10, 5 is sum of two consecutive primes,
2 and 3. 5 is the prime number which is sum of largest
chain of consecutive below limit 10.

Below limit 30, 17 is sum of four consecutive primes.
2 + 3 + 5 + 7 = 17
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Below are steps.

1. Find all prime numbers below a maximum limit (10^6) using Sieve of Sundaram and store them in primes[].
2. Construct a prefix sum array prime_sum[] for all prime numbers in primes[]
prime_sum[i+1] = prime_sum[i] + primes[i].
Difference between two values in prime_sum[i] and prime_sum[j] represents sum of consecutive primes from index i to index j.
3. Traverse two loops , outer loop from i (0 to limit) and inner loop from j (0 to i)
4. For every i, inner loop traverse (0 to i), we check if current sum of consecutive primes (consSum = prime_sum[i] – prime_sum[j]) is prime number or not (we search consSum in prime[] using Binary search).
5. If consSum is prime number then we update the result if the current length is more than length of current result.

Below is implementation of above steps.

```// C++ program to find Longest Sum of consecutive
// primes
#include<bits/stdc++.h>
using namespace std;
const int MAX  = 10000;

// utility function for sieve of sundaram
void sieveSundaram(vector <int> &primes)
{
// In general Sieve of Sundaram, produces primes smaller
// than (2*x + 2) for a number given number x. Since
// we want primes smaller than MAX, we reduce MAX to half
// This array is used to separate numbers of the form
// i+j+2ij from others where 1 <= i <= j
bool marked[MAX/2 + 1] = {0};

// Main logic of Sundaram. Mark all numbers which
// do not generate prime number by doing 2*i+1
for (int i=1; i<=(sqrt(MAX)-1)/2; i++)
for (int j=(i*(i+1))<<1; j<=MAX/2; j=j+2*i+1)
marked[j] = true;

// Since 2 is a prime number
primes.push_back(2);

// Print other primes. Remaining primes are of the
// form 2*i + 1 such that marked[i] is false.
for (int i=1; i<=MAX/2; i++)
if (marked[i] == false)
primes.push_back(2*i + 1);
}

// function find the prime number which can be written
// as the sum of the most consecutive primes
int LSCPUtil(int limit, vector<int> &prime, long long int sum_prime[])
{
// To store maximum length of consecutive primes that can
// sum to a limit
int max_length = -1;

// The prime number (or result) that can be reprsented as
// sum of maximum number of primes.
int prime_number = -1;

// Conisder all lengths of consecutive primes below limit.
for (int i=0; prime[i]<=limit; i++)
{
for (int j=0; j<i; j++)
{
// if we cross the limit, then break the loop
if (sum_prime[i] - sum_prime[j] > limit)
break;

// sum_prime[i]-sum_prime[j] is prime number or not
long long int consSum  = sum_prime[i] - sum_prime[j];

// Check if sum of current length of consecutives is
// prime or not.
if (binary_search(prime.begin(), prime.end(), consSum))
{
// update the length and prime number
if (max_length < i-j+1)
{
max_length = i-j+1;
prime_number = consSum;
}
}
}
}

return prime_number;
}

// Returns the prime number that can written as sum
// of longest chain of consecutive primes.
void LSCP(int arr[], int n)
{
// Store prime number in vector
vector<int> primes;
sieveSundaram(primes);

long long int sum_prime[primes.size() + 1];

// Calculate sum of prime numbers and store them
// in sum_prime array. sum_prime[i] stores sum of
// prime numbers from primes[0] to primes[i-1]
sum_prime[0] = 0;
for (int i = 1 ; i <= primes.size(); i++)
sum_prime[i] = primes[i-1] + sum_prime[i-1];

// Process all queries one by one
for (int i=0; i<n; i++)
cout << LSCPUtil(arr[i], primes, sum_prime) << " ";
}

// Driver program
int main()
{
int arr[] = {10, 30, 40, 50, 1000};
int n = sizeof(arr)/sizeof(arr[0]);
LSCP(arr, n);
return 0;
}
```

Output:

```5 17 17 41 953
```

This article is contributed by Nishant_singh (pintu). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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