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