Author: Aryan KD
Tester: Aryan KD
Editorialist: Aryan KD
DIFFICULTY:
CAKEWALK, SIMPLE, EASY.
PREREQUISITES:
Math .
PROBLEM:
Series of even natural number is given . 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 …… 10^9
you have to find a missing number if sum of first N number is given
for example :
N=5, Sum=24
you have to find missing term in first 5 number
according to above series sum of first 5 number is 30, and if we subtract 6 from it then sum becomes 24 hence the missing number is 6 .
EXPLANATION:
You simply given N and sum of first N even number in that one term is missing so if we subtract the sum given in question with original sum that given we will get our answer.
for example 
N=5 sum=25
Sum of first N even number is 30 but sum is given 25 so if we subtract 25 with 30 we will get our answer that is 5
formula for sum of first N even natural number is (N*(N+1))
SOLUTIONS:
Setter's Solution
#include<bits/stdc++.h>
using namespace std;
void testcase()
{
long long int n,sum;
cin>>n>>sum;
long long int k=n*(n+1);
long long int pp=k-sum;
cout<<pp;
}
int main()
{
int t;
cin>>t;
while(t–)
{
testcase();
cout<<endl;
}
return 0;
}
Tester's Solution
#include<bits/stdc++.h>
using namespace std;
void testcase()
{
long long int n,sum;
cin>>n>>sum;
long long int k=n*(n+1);
long long int pp=k-sum;
cout<<pp;
}
int main()
{
int t;
cin>>t;
while(t–)
{
testcase();
cout<<endl;
}
return 0;
}
Editorialist's Solution
#include<bits/stdc++.h>
using namespace std;
void testcase()
{
long long int n,sum;
cin>>n>>sum;
long long int k=n*(n+1);
long long int pp=k-sum;
cout<<pp;
}
int main()
{
int t;
cin>>t;
while(t–)
{
testcase();
cout<<endl;
}
return 0;
}
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