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

Below are steps.

- Find all prime numbers below a maximum limit (10^6) using Sieve of Sundaram and store them in primes[].
- 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. - Traverse two loops , outer loop from i (0 to limit) and inner loop from j (0 to i)
- 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). - 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:

1 17 17 41 953