# UTMOPR - Editorial

Author: Ankit Srivastava and Uttam Kanodia

Tester: Roman Rubanenko
Editorialist: Amit Pandey

Cakewalk

### PREREQUISITES:

Basic programming

### PROBLEM:

Given an array of n integers, you have to do the following operation K times: Insert a number in the array which is strictly greater than current sum of the array. You have to find whether the last number inserted is even or odd.

### QUICK EXPLANATION:

As the value of K is small, we can keep inserting smallest possible numbers at each step and print the K^{th} number modulo 2.

### EXPLANATION:

This problem can be solved by simulating the operation given in the problem.

First find the sum of the given array(A) . This can be done in O(N) time using a single loop. As we have to print the K^{th} element to be inserted (modulo 2), we can replace S by S\ \% \ 2.

S = 0;
for(int i=0; i< n; i++)
S = S + A[i];
S = S%2;


Now make an array B of size (K+1), B_i will denote the i^{th} element to be inserted, and B_K will be our required answer. At any step, if parity of the sum of the elements of array is “even”, parity of inserted element will be “odd”.

B[1] = (S + 1) % 2 ;   // we will insert the smallest possible number at each step
total_sum = 0;
for(int i=2 ; i< K ; i++)
{
total_sum = (total_sum + B[i-1]);
B[i] = (total_sum + 1) % 2; // insert a number greater than current sum of the array.
}


Finally we can output B_K. The time complexity of the whole code will be \mathcal{O}(N + K).

Another solution
Let us say that sum of the array A is even. The next inserted element will be odd, now sum of array will be odd, so next inserted element will be even, now sum of array becomes odd, so we will insert an even number, and so on. So we can generalize that if sum of array is even, then for K = 1, last inserted number will be odd, otherwise it will be even.

Now, we will consider the case in which sum of the array A is odd. The next inserted element will be even, now sum of array will become odd, so next inserted element will be even, now sum of array will be odd, we will add another even number, and so on. So we can generalize that last inserted number is always even in this case.

So finally, we can obtain the following simple solution.

Compute sum of array.
If K = 1:
if sum is even:
print "odd"
else:
print "even"
else:
print "even"


### Solution:

Setter’s solution can be found here
Tester’s solution can be found here

3 Likes

We can also observe a pattern forming in the array. First, the sum of all elements in the array is computed to be S.
Then, the array fans out as: S, S+1, 2S+2, 4S+4, 8S+8, 16S+16…

The kth term is equal to 2^(k-1)*(S+1)

So our answer is odd if and only if S is even and K is equal to 1. Complexity: O(N).

1 Like

because Tester and Setter solution not yet updated …

here is my code…

#include<bits/stdc++.h>
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t,n,k;
int sum=0,val=0,ans=0;
scanf("%d",&t);

while(t--){

scanf("%d%d",&n,&k);
sum=0;

while(n--){
scanf("%d",&val);
sum+=val;
}
if(sum%2==0){

if(k==1) printf("odd\n");
else printf("even\n");
}
else printf("even\n");
}
return 0;


}

End of code…HAPPY CODING

2 Likes

No array will be not like S, S+1, 2S+2, 4S+4, 8S+8, 16S+16.
It will be like S, 2S+1, 4S+3, 8S+7, 16S+15…

kth term -> S * ( 2 ^ ( k - 1 ) ) + ( 2 ^ ( k - 1 ) - 1 ).

Then we can get answer by putting values of S and k.

1 Like

@amitpandeykgp…how to paste code with indentation here…

@rcsldav2017 press CTRL+K and start typing your code

2 Likes

I wonder how could this solution be accepted! It just prints “odd” if k equals 1, and prints “even” otherwise.

2 Likes

Here is my solution too short
#include<bits/stdc++.h>
using namespace std;
int main()
{
long long int t,n,k,i,s,a,p;
cin>>t;
while(t–)
{s=0;
cin>>n>>k;
for(i=0;i<n;i++)
{
cin>>a;
s+=a;
}
p=(s+1)*(long long int)pow(2,k-1);
if(p%2==0)
cout<<“even”<<endl;
else
cout<<“odd”<<endl;
}
}

Can someone point out the mistake in my solution?

I guess it’s pretty much same as @rcsldav2017 's solution.


#include <bits/stdc++.h>
using namespace std;
int main()
{
ios_base::sync_with_stdio(false);
int n,k,x,s=0;
cin>>n>>k;
while(n--)
{
cin>>x;
s += x%2;
}
if(k==1)
{
if(s%2)
cout<<"even"<<endl;
else
cout<<"odd"<<endl;
}
else
cout<<"even"<<endl;
return 0;
}


1 Like

@nikhilhassija

Number of test cases.
Your code should run for Various test case T.
You are not scanning that.

1 Like

@nikhilhassija

hi…

actually your logic is same as mine and

• your logic is right too

• what is wrong with your code

• why u got WA?

here what you have missed…

just look at constraints

• Constraints:

• 1 ≤ T ≤10
• 1 ≤ K ≤ pow(10,6)
• 1 ≤ N ≤ pow(10,3)
• 1 ≤ array element ≤ pow(10,9)

-Hope you get your mistake …still not…No problem

MISTAKE

• 1<=array element<pow(10,9)
• what You have done here… int n,k,x,s=0;
• ***how can u store x[1,pow(10,9)] into integer data type ***
• *** Your are thinking here is no test case …still believe look at constraints again…):P***

Solution

• *** make x and sum variable long or long long***

-*** there are multiple test cases… look at 1<=T<=10***

here is your corrected code…still did not get it …click on me

for more conversation join my group…Novice Programmers @ NITK

## Happy coding

#include<stdio.h>
#include<stdlib.h>

int main()
{
int t,n,k,j,i,sum=0;

scanf("%d",&t);
for(i=0;i<t;i++)
{

scanf("%d%d",&n,&k);
int *N=(int *)malloc((n+k)*sizeof(int));
for(i=0;i<n;i++)
{
scanf("%d",N+i);
sum=sum+*(N+i);

}
if(k==1)
{
if(sum%2==0)
printf("odd");
else
printf("even");

}
else
printf("even");

}


}

public static String logicHere(String fileName){
int opArr[];
int N=0;
int K=0;
String line2="";
String line3="";
int initialSum=0;
String retValue="";
try {
String[] tmpCh = line2.split(" “);
N=Integer.parseInt(tmpCh[0]);
K=Integer.parseInt(tmpCh[1]);
opArr = new int[N+K];
tmpCh = line3.split(” ");
for(int i=0;i<N;i++){
int tmp=Integer.parseInt(tmpCh[i]);
opArr[i]= tmp%10;
initialSum=(initialSum + tmp)%10;
}
for(int i=N;i<K;i++){
opArr[i]=(initialSum%10)+1;
initialSum=(initialSum+opArr[i])%10;
}
if(opArr[K]%2==0)
retValue=“even”;
else
retValue=“odd”;

	}
catch(Exception ex){
ex.printStackTrace();
}
return retValue;
}

Why everyone is trying to take an array and then evaluate the sum. There’s no need of it at all.
We just need to look for number of odd elements in the array. if number of odd elements is even then sum is always even.if number of odd elements is odd them sum is always odd.

enter code here
import java.util.*;


import java.math.*;
import java.util.regex.Pattern;
public class check {

// Driver code
public static void main(String[] args)
{


Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t>0){
int n=sc.nextInt();
int k=sc.nextInt();
int count=0;
for(int i=0;i<n;i++){
if(sc.nextInt()%2!=0){
count++;
}
}
if(count%2==0){
if(k==1)
System.out.println(“odd”);
else
System.out.println(“even”);
}else{

    System.out.println("even");


}
t–;
}

}


}

#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
while(t–)
{
int n,k;
cin>>n>>k;
long long s=0,a[n+5];
for(int i=0;i<n;i++)
{
cin>>a[i];
s+=a[i];
}
if(k==1)
cout<<((s+1)%2?“odd”:“even”);
else if(k==2)
cout<<((2*s+1)%2?“odd”:“even”);
else
cout<<“even”;
cout<<endl;
}
return 0;
}

Plz can someone point out mistake in my code plz.Here is my code:
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
while(t–)
{
long long int n,k,i,s=0;
cin>>n>>k;
long long int a[n];
for(i=0;i<n;i++){

	cin>>a[i];
if(a[i]%2==1)
s++;

}
if(s%2==1)
s=1;
else
s=0;
if(k==1)
{
if(s==1)
cout<<"odd"<<endl;
else
cout<<"even"<<endl;
}
else
cout<<"even"<<endl;
}
return 0;


}

#include<bits/stdc++.h>
using namespace std;

int main()
{
int t;
cin>>t;

while(t--)
{
int n,k;
cin>>n>>k;
long int a[n],s=0;
for(int i=0;i<n;i++)
{
cin>>a[i];
s+=a[i];
}

if(s%2==0&&k==1)
{
cout<<"odd\n";
}
else
cout<<"even\n";
}
return 0;


}

lol, thats his luck

@snk967:-
That happened because of some weak test cases at corner!

@shubham99 … THANKS DUDE…IT WORKS