C++ Program for Consecutive Prime Sum (TCS Codevita) | PrepInsta

Cpp program for consecutive prime sum

Consecutive Prime Sum Problem

TCS CodeVita is one of the most prestigious coding competition that is held on a global platform by TCS in the search of world’s best coder. It is being conducted from last 8 years now. Consecutive prime sum is one of the sample problem for TCS CodeVita season 9 edition. Here we have provided a C++ solution for Consecutive Prime Sum problem

Problem Description

Question – :  Some prime numbers can be expressed as a 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 the number of prime numbers that satisfy the above-mentioned property in a given range.

Input Format: First line contains a number N

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

Constraints: 2<N<=12,000,000,000

S.no Input Output Comment
1 20 2 (Below 20 there are two such members; 5 and 17) 5=2+3 17=2+3+65+7
2 15 1

Cpp Code for Consecutive Prime Sum Problem

#include <iostream>
using namespace std;

int prime(int b)
{
    int j,cnt;
    cnt=1;
     for(j=2;j<=b/2;j++)
     {
         if(b%j==0)
         cnt=0;
     }
     if(cnt==0)
     return 1;
     else
     return 0;
}
int main() 
{
 int i,j,n,cnt,a[25],c,sum=0,count=0,k=0;
 cout<<"Enter a number : "; 
cin>>n;
 for(i=2;i<=n;i++)
 {
     cnt=1;
     for(j=2;j<=n/2;j++)
     {
         if(i%j==0)
         cnt=0;
     }
     if(cnt==1)
     {
        a[k]=i;
        k++;
        }
 }
 for(i=0;i<k;i++)
 {
     sum=sum+a[i];
    c= prime(sum);
    if(c==1)
    count++;
 }
 cout<<count;
 return 0;
}
Output
Enter a number : 5
1

Consecutive Prime Sum Problem in Other Coding Languages

C

To find the solution of Consecutive Prime sum problem in C Programming language click on the button below:

C

Java

To find the solution of Consecutive Prime sum problem in Java Programming language click on the button below:

Java

Python

To find the solution of Consecutive Prime sum problem in Python Programming language click on the button below:

 Python

8 comments on “C++ Program for Consecutive Prime Sum (TCS Codevita) | PrepInsta”


  • Shobhit

    solved with help of set and vector
    first find prime numbers upto n and store it in vector
    second – make a loop for elements of vector which contains prime numbers
    third – make a variable sum and store it in set
    and check with the help of if statement

    #include
    using namespace std;

    int main()
    {
    int n;
    cin>>n;
    set s;
    int sum = 0;
    int ans = 0;
    vector vec;
    vec.emplace_back(2);
    for(int i=3;i<=n;i++){
    int count=0;
    for(int j=2;j<i;j++){
    if(i%j==0){
    count++;
    break;
    }
    }
    if(count==0){
    vec.emplace_back(i);
    }
    }
    for(int i=0;i<vec.size();i++){
    sum += vec[i];
    if(s.find(vec[i])!=s.end()){
    ans++;
    }
    s.emplace(sum);
    }
    cout<<ans;

    }


  • 18EBKCS112_VAIBHAV_POLIWAL

    We can solve this problem using Sieve of Eratosthenes… here is my solution in c++
    #include
    #include
    using namespace std;

    int main(){
    int N;
    cin>>N;

    bool primeArray[N] = {0};
    for(int i=2;i<N;i++){
    if(primeArray[i]==0){
    for(int j=i*i;j<N;j+=i){
    primeArray[j] = 1;
    }
    }
    }
    primeArray[0] = 1;
    primeArray[1] = 1;
    int sum = 0;
    vector vi;
    /*for(int i=2;i<N;i++){
    if(primeArray[i]==0 && sum<=N){
    sum += i;
    if(primeArray[sum]==0 && sum <= N){
    vi.push_back(sum);
    }
    }
    }*/ // same loop using while loop
    int i=2;
    while(i<N){
    if(primeArray[i]==0){
    sum += i;
    if(primeArray[sum]==0 && sum<=N && sum!=2){
    vi.push_back(sum);
    }
    }
    i++;
    }
    cout<<"Size is : ";
    cout<<vi.size()<<"\n"; // this is the answer
    for(int j=0;j<vi.size();j++){ // this will print all that number
    cout<<vi[j]<<" ";
    }
    return 0;
    }

    Time Complexity = O(N Log N)
    Space Complexity = O(N)


  • Akshay

    #include

    using namespace std;
    int prime(int p)
    {
    int count=0;
    for(int j=2;j>n;
    for(i=2;i<=n;i++)
    {
    s=prime(i);
    if(s==0)
    {
    //cout<<i<<" ";
    sum=sum+i;
    //cout<<sum<2 && sum<n)
    {
    //cout<<sum<<" ";
    k++;
    }
    }
    }
    cout<<k;

    return 0;
    }


  • Omkar

    #include
    #include
    using namespace std;

    int main()
    {
    vector p;
    int n;
    cin>>n;
    for(int i=2;i<=n;i++)
    {
    bool prime=true;
    for(int j=2;j<i;j++)
    {
    if(i%j==0)
    {
    prime=false;
    break;
    }
    else
    prime=true;
    }
    if(prime==true)
    p.push_back(i);
    }
    int count=0;
    for(int i=2;i<p.size();i++)
    {
    int sum=0;
    for(int j=0;jp.at(i))
    break;
    }
    }
    cout<<count;
    }


    • Jaswanth Reddy

      #include
      using namespace std;
      int main(){
      int num,count=0;
      cin>>num;
      for(int i=2;i<=num/2;i++){
      int sec=num-i,flag=1,flag1=1;
      for(int k=2;k<=i/2;k++){
      if(i%k==0){
      flag=0;
      }
      }
      if(flag==1){
      for(int j=2;j<=sec/2;j++){
      if(sec%j==0){
      flag=0;
      }
      }
      }
      if(flag==1 && flag1==1){
      ++count;
      }
      }
      cout<<count;
      }


    • HelpPrepInsta

      Yes Manish, we are fixing it, if you have the correct answer for this problem, you can submit your answer in the comment section only, and we’ll post it on our page