CLPERM - Editorial

this link will help you

1 Like

I did this problem using the same idea. However I got WA in the last 2 subtasks.
int n , k;
cin >> n >> k;

	vector<ll> temp(k);
	rep( i , 0 , k )cin >>temp[i];
	sort( all( temp ) );
	ll x = ( n*( n + 1 ) ) / 2 ;
	ll last = 0;
	for ( ll a : temp )
	{
		last += a;
		x = ((a)*( a + 1 )) / 2;
		x -= last;
		if ( a > x )break;
	}
	++x;
	if ( x % 2 )printf( "Chef" );
	else printf( "Mom" );
	printf( "\n" );

What’s wrong here ?

Hello everyone,

Here is the author’s commented solution for this problem.

Hope you find it useful …

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I couldn’t solve the problem in the competition and read the tutorial so as to implement its solution. Although, I had some difficulty in understanding the solution, I was able to write its code. The solution fails and gives WA for task # 0,1,2 and 5. What could be the bug in my code?

CODE LINK

Thanks.

Regards,
Ankit.

Hello ankit,

I checked your solution just now and find that your logic is all correct except for the part that you are not taking care of overflow that will surely occur during the third subtask. Use of long long instead of int will solve this problem.

http://www.codechef.com/viewsolution/5901687

Here is the Ac version of your solution. I just got Ac with your code :slight_smile: :B

Hope you find my words useful to you

@ma5termind, yea I missed considering the case of having large numbers.
Thanks.

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How can we be sure that if (Bi)+1 to B(i+1)-1 are available then M+1 to M+S are possible

from the time i first solved this problem to this date, i am not able to understand why am i getting run time error, when i try to delete those dynamic arrays, i get wrong answers, really need ur help coders.here is my solution CodeChef: Practical coding for everyone

@ashrko619: here is AC submission of your code. You missed the case that, if your for loop is executed fully, updated x will be “sum(1 to N)-last”. Hope it helps

Can we have someone to proofread the editorials please?! Use of ambiguous English leads to incomprehensible logic no matter how easy it is!

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I think to find maximum unachievable sum, instead of keeping track of range between every missing number

we can do

SumOfAll(natural)Numbers upto (Present missing number - 1) - Sum of all previous missing numbers.

e.g 3 7 10 are missing numbers

to find sum at missing number 10
we can find 9 * (9+1)/2 - (3 + 7)

Is it correct

IN The basic O(N) solution for the problem that given a sorted array of no.s,find the smallest no. which cannot be formed by any subset of no.s from the array.

If we asssume that the smallest unachievable sum is res till arr[i-1] then it is the answer if arr[i]>res
That part is quite clear…
But the fact that :
the new res = res + arr[i] if arr[i] <= res is not clearly proven… How can we say that the new res cannot also be formed by some elements from index 0…i-1 from the array…
If we can say that it cannot be achieved by any subset from 0…i-1 then it is definitely the new res…

Can anybody explain that part with some proof.

This

@darkshadows:
Links to solution don’t work.
Also, editorialist’s English is shitty.
:frowning:

#include
#include
using namespace std;

int main()
{
std::cout.sync_with_stdio(false);
unsigned long long int t;
cin>>t;

while(t--)
{
   unsigned long long int n,k;
   cin>>n>>k;
   unsigned long long int res=1;

        unsigned long long int a[k],b[n-k];
       for(unsigned long long int i=0;i<k;i++) cin>>a[i];
       sort(a,a+k);

       unsigned long long int i=1,j=0,x=0;

       while(i<=n)
       {
           if(a[j]==i)
              {
                i++;
                j++;
              }

           else
           {
               b[x]=i;
               i++;
               x++;
           }
       }
       sort(b,b+n-k);
       for(i=0;i<n-k&&b[i]<=res;i++)
        res+=b[i];
    //cout<<res<<endl;
    if(res%2==0)
        cout<<"Mom"<<endl;
    else
        cout<<"Chef"<<endl;

}
return 0;

}

I am getting run-time error(SIGSEGV) in task 3 only can you tell me why?

why i am getting WA for sub tasks 1 and 3 here is my code link CodeChef: Practical coding for everyone

You would have to sort the array B any ways so even if you use count sort you cannot have a complexity better than O(N)

try reading the link given for better clarity at the last of the editorial. basically at any moment if you can form sum (s) using numbers preceding a number (x), but (x > s+1), then you can’t form (sum+1). As, by using the previous numbers, you can only form sum (s), and if you take (x), does not matter which number you add to it (or even if you don’t add any), it will always be greater than (s+1), and hence you will not be able to form (s+1)

What that means is, if we are given numbers in the range [1 to i], then we can form all numbers in the range [1,i*(i+1)/2]. So, for every b[i], we calculate the sum of (1,b[i]-1). Let the maximum number we can create from these numbers is M. The next number we want is M+1, which we can create using b[i]. If b[i+1]> M+1, then it’s impossible to create it and thus ans would be M+1.

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Will the links to setter’s and tester’s solutions be updated ever?