Problem : (CodeChef: Practical coding for everyone)
My Solution is given TLE.
My Solution:

``````#include<bits/stdc++.h>
#include<algorithm>
#define fast ios_base::sync_with_stdio(false);cin.tie(0);
#define lli long long int
#define vi vector<lli>
#define pb push_back
#define fr(n) for(lli i=0;i<n;i++)
#include<string>
#define mp map<char,lli>
#define umap map<lli,lli>
#define test lli t;cin>>t;while(t--)
using namespace std;
int main()
{
test
{
lli l,r;
cin>>l>>r;
lli sum=0;
for(lli i=l;i<=r;i++)
sum+=i;
lli total=0;
for(lli i=l;i<=r;i++)
{
sum-=i;
total+=(sum*i);
}
cout<<total*2<<endl;
}
}
``````

Formulate what you are doing.

You can think of effective approach from here

You must get an idea after seeing the constraints that you simply cannot use a loop here, and also these are some of the basic algebraic formulas I think you must have used many number of times,
and I think thatâ€™s why even the question is named so.

Solution:

``````#include<bits/stdc++.h>
#include<algorithm>
#define fast ios_base::sync_with_stdio(false);cin.tie(0);
#define lli long long int
#define vi vector<lli>
#define pb push_back
#include<string>
#define mp map<char,lli>
#define umap map<lli,lli>
#define test lli t;cin>>t;while(t--)
#define mod 1000000007
using namespace std;
int main()
{
lli l,r;
cin>>l>>r;
lli sum=0;
for(lli i=l;i<=r;i++)
sum+=i;
lli val=sum*sum;
lli total=0;
for(lli i=l;i<=r;i++)
total+=(i*i);
cout<<(val-total)%mod<<endl;
}
``````

There are two things missing here :-
1. A loop for testcases
2. Overflow error