Author: Aryan KD
Tester: Aryan KD
Editorialist: Aryan KD
DIFFICULTY:
CAKEWALK, SIMPLE, EASY.
PREREQUISITES:
Math .
PROBLEM:
Noob is very intelligent in his class , he always play many mind games there for he has sharp mind but now his teacher is given a challenge to him he is given a series and he has to find Nth term in the series.
0, 0, 2, 1, 4, 2, 6, 3, 8, 4, 10, 5, 12 . . .
EXPLANATION:
We have to find Nth term in given sequence here if we observe then there is in odd index all the number is addition by 2 in previous number and in even place all the number is addition of 1 in previous number so we have to iterate our loop according to given condition and find out our term
suppose N = 5
here N is 5 that is odd number there for we have to find addition of 2 in two times hence the output is 4
SOLUTIONS:
Setter's Solution
#include<bits/stdc++.h>
using namespace std;
void testcase()
{ //0, 0, 2, 1, 4, 2, 6, 3, 8, 4, 10, 5, 12 . . .
long long int n,k=0,sum=0;
cin>>n;
if(n<3){
cout<<“0”;
}
else if(n%2!=0){
n=n/2+1;
k=(n-1)*2;
cout<<k;
}
else{
n=n/2;
sum=(n-1);
cout<<sum;
}
}
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()
{ //0, 0, 2, 1, 4, 2, 6, 3, 8, 4, 10, 5, 12 . . .
long long int n,k=0,sum=0;
cin>>n;
if(n<3){
cout<<“0”;
}
else if(n%2!=0){
n=n/2+1;
k=(n-1)*2;
cout<<k;
}
else{
n=n/2;
sum=(n-1);
cout<<sum;
}
}
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()
{ //0, 0, 2, 1, 4, 2, 6, 3, 8, 4, 10, 5, 12 . . .
long long int n,k=0,sum=0;
cin>>n;
if(n<3){
cout<<“0”;
}
else if(n%2!=0){
n=n/2+1;
k=(n-1)*2;
cout<<k;
}
else{
n=n/2;
sum=(n-1);
cout<<sum;
}
}
int main()
{
int t;
cin>>t;
while(t–)
{
testcase();
cout<<endl;
}
return 0;
}