**CRACK1**: Editorial

**Problem-code**: CRACK1

**Contest-Code**: CSMH2021

**Author:** RAPETI MAHENDRA

**Editorialist:** RAPETI MAHENDRA

# DIFFICULTY:

Simple

# PREREQUISITES:

Basic observations, Sieve of Eratosthenes

# PROBLEM:

One day Chef teacher gives him a problem to solve.

The problem is as follows:

Given a number N , print the number corresponding to input by observing following series

## 0 2 5 5 10 10 ...

Chef took it as a challange but struggling a lot to find series help him carefully so that his teacher don't notice you!!# EXPLANATION

on observing the series it is easy to conclude that for given n the answer is just sum of all prime numbers ranging from 1 to n.

As n is very small we can simply check for primes (or) use sieve of eratosthenes.

# SOLUTION:

#include<bits/stdc++.h>

using namespace std;

int main()

{

int t,n,i,j;

cin>>t;

bool prime[n+1];

memset(prime,true,sizeof(prime));

for(i=2;i*i<=n;i++)
{
if(prime[i]==true)
{
for(j=i*i;j<=n;)

{

prime[j]=false;

j+=i;

}

}

```
}
while(t--)
{
cin>>n;
int sum=0;
for(i=2;i<=n;i++)
{
if(prime[i])
{
sum+=i;
}
}
cout<<sum<<endl;
}
return 0;
```

}

Feel free to Share your approach, if you want to. (even if its same ) . Suggestions are welcomed as always had been.