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Hello Codeforces,

I am happy to invite you to GeeksforGeeks's Pan-India Coding Contest – Code India Code. Showcase your coding skills and win rewards up to INR 7 Lakhs.

Contest Details are as follows:

Qualification Round: (Register Here)

Sunday, 6th March 2022, 12:00 AM (IST) to 7th March 2022,12:00 AM (IST)
In order to qualify in the first round, the participant needs to solve a minimum of 1 question out of 4 in 24 hours.

Final Round:

Thursday, 10 March 2022, 7:30 PM to 10:00 PM IST
Only those who qualify in the first round will receive a link to the final round.

Prizes:

Rank 1: Macbook Pro + Course voucher worth INR 12,000
Rank 2: iPad Air + Course voucher worth INR 7,000
Rank 3: Smart Watch + Course Voucher worth INR 3,000
Rank 4-25: Tablets + Course Voucher upto INR 1500
Rank 26-100: Noise Earbuds + Course Voucher upto INR 1500
Rank 101-200: Noise Smartbands + Course Voucher upto INR 1500
Prize distribution is limited to those residing in India only.

Problem Setting Panel:

Contest Coordinators: Ashishgup, AvinashKartik
Problem Setters: Ashishgup, AvinashKartik, Utkarsh.25dec, bunty_x, i.am.pratik, Omja
Problem Tester: manpreet.singh

Even though the prizes are only for Indians, I would like to invite others to participate — the problem-set (specially the finals) contains some interesting and challenging problems :D

Good luck everyone. See you on the leaderboard!

pos [0,19] -> position from left we are at small[0,1] -> are we smaller than the input number or equal to it till the i position start[0,1] -> have we started building the number zero[0,1] -> is the product of digits equal to 0 sumg[0,2] -> is the sum of all quartic power of digits greater than 1 (helpful in handling cases like 1000) two[0,1] -> does the prime factorization of product of digit have 2 as a prime factor three[0,1] -> does the prime factorization of product of digit have 3 as a prime factor five[0,1] -> does the prime factorization of product of digit have 5 as a prime factor seven[0,1] -> does the prime factorization of product of digit have 7 as a prime factor mod2[0,1] -> the mod of the sum of the quartic power of digits with 2 mod3[0,2] -> the mod of the sum of the quartic power of digits with 3 mod5[0,4] -> the mod of the sum of the quartic power of digits with 5 mod7[0,6] -> the mod of the sum of the quartic power of digits with 7

void solve(){ int a, b; cin >> a >> b; if(b <= a){ printf("%d\n", (a-b)); return; } if((a|b) == b || a + 1 == b){ printf("1\n"); return; } bool i1 = ((a+1)|b) == (b); bool i2 = (a|(b+1)) == (b+1); if(i1 || i2 || (a + 2 == b)){ printf("2\n"); return; } printf("3\n"); }

Комментарии (48)

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Ashishgup

3 года назад, # |

Auto comment: topic has been updated by Ashishgup (previous revision, new revision, compare).

→ Ответить

prabhav_

Cool prizes!

A-reum

reply challenge on same day

kal013

Will the qualification round results be considered in any way while deciding the rankings for the prizes?

bunty_x

3 года назад, # ^ |

No :)

ilovecheapthrills

+43

Congrats on winning Macbook Pro!!

Aabhas99

+104

Top performers will be given course vouchers, but logically thinking, do the top performers need those courses???

amul02

← Rev. 2 →

Right. I think they should provide the course voucher to those who perform average in the contest. So that they can enhance their skill with their expertise.

nor

+19

Is using prewritten code allowed? The rules don't seem to address this.

Omja

Yes :)

Really Excited to participate.

towrist

+13

are the prizes be for Indians in top-X, or top-X Indian contestants in a separate Indian ranklist?

Top X Indian contestants

disredoc

+18

Is testcase of "MINIMUM MOVES" is wrong?

+31

Test cases are correct.

he1senb3rg

will we enter the ranklist only after clicking on finish test? Edit :No

cheatreport

+28

Mass cheating is going on!

A youtube channel is revealing the solution for problem A and problem B.

Please ban the youtube channel as soon as possible :(:(:(

18o3

Please remove the links.

you could have told this in private to any of the organisers.

ALuca_Rd

Can anybody share the solution of the last 3 problems.

+26

Don't know if it's intended or not but I solved Problem D(Special Number Count) with Digit Dp.

My dp solution has $$$13$$$ states which are as follows:

In all the memory declaration for my dp looked like this

ll dp[20][2][2][2][2][2][2][3][5][7][2][3][2];

The recursive function looked like this

ll recursion(int pos,int small,int two,int three,int five,int seven,int mod2,int mod3,int mod5,int mod7, int zero , int sumg1 ,int start ,string &curr)

Each recursive call was somewhat like this

val+=recursion(pos+1,0,n2,n3,n5,n7,(mod2+k)%2,(mod3+k)%3,(mod5+k)%5,(mod7+k)%7,n0,nsumg1,0,curr);

My somewhat okayish implementation

Heartfelt sincere orz!! 13 state DP

gupta_samarth

I did the same basically, but you don't really need $$$13$$$ states. You can compress quartic sum modulo each prime into a single value upto $$$210$$$ ( using custom radix base $$$\{2,3,5,7\}$$$ ). Also, similarly you can compress whether product contains $$$2/3/5/7$$$ into a 4 digit bitmask.

So my dp looked like dp[20][210][1<<4][2][2][2]

Nice solution.

codemastercpp

+35

Downvoted for having the problem A in the contest

tiasmondal

Can anybody share the solution to the first problem :)). Tried everything.

← Rev. 7 →

It's similar to this Problem. The only difference I found is that you don't need more than 3 operations to convert A into B.

But the solution is entirely different, and also here update operation is A = A | X where X could be any number >= 0

kasparovian

+16

Basically, if B <= A, we need to do |B-A| moves, as we can only increase B by 1. Otherwise, it is not difficult to see that we can always do it in at most 3 moves, as we will take the highest bit (say i) that they differ at, and in 1 move set all the bits in A smaller than i (by or-ing) and then increment A by 1, and then or the remaining bits smaller than i to make A and B equal. Now that we know the answer is always <= min(B-A, 3), we only need to check for answers 1 and 2. This is easy, as we only need to check for 4 cases, (A|X) = B, ((A+1)|X) = B, (A|X) = B+1, (A|X)+1 = B. For the last case, just reduce B by 1 and check.

shreyash1488

Hey, this implementation handles all the cases you described. Why its giving WA?

I don't think you have handled the last case, (A|X) + 1 = B

Thanks got it.

This case just took my entire SUNDAY.

If B > A

Somehow the main ambition is to make A a sub mask of B. Now, if A is already a sub mask of B, we can make them the same in only one operation!

let i be the most significant bit where A and B are not the same

update A = A | (2 ^ i - 1)

for example A = 10101 and B = 11010, we update A = A | 7, now A becomes 10111

update A = A + 1

Now A becomes the sub mask of B for sure, and you will need only one operation to make both equal!

So, at most, three operations are required to make A and B equal.

We have to handle cases where only one or two operations are sufficient.

Thanks for the detailed explanation

-is-this-fft-

+207

Am I the only one that is always bothered by statements like "win rewards up to INR 7 Lakhs"? Because it is not actually possible to win 7 lakh; the best you can do is a MacBook (which I can't imagine costing 7 lakh). 7 lakh appears to be the total value of the rewards; in addition, none of it is actually in the form of money, which is even more false advertising. Also, about 2 lakh comes from GFG course vouchers, which I suspect aren't really worth anything.

Scythe

+56

Also, about 2 lakh comes from GFG course vouchers, which I suspect aren't really worth anything.

Destroyed in seconds

+10

GFG course vouchers are not only limited to programming/dsa courses, you can take any( OS, DBMS, etc.) course.

+66

He is still likely correct.

but when did he said he is wrong?

gotcoding124

Yeah you are correct it is rather a way of advertising it to suit their needs for greater participation

surya535

Can someone explain how to solve problem — The Duality Task?

Electron

You can find duality by considering a common prime number for each element in a subsequence. Now, just take initial C as B and find its duality. ans = n — max(0, duality_A — duality_B)

ikr

How to solve B — “Split the Array”? I had nothing better than an $$$N^2$$$ DP…

It involved a bit of casework.

Think along the lines of what will happen if k is odd/even and j is odd/even.

The problem then reduced to a dp with $$$4$$$ states per element.

aryanc403

← Rev. 5 →

That $$$O(n^2)$$$ is enough just to ensure that you don't create a segment of length more than 20. So it will become $$$O(20*N)$$$

You can work on paper to see that if a segment has a length of more than 10 then you can split the first/last 4 elements into 2 segments of length 2 and it won't change signs. These 10/20 are the loose upper bounds, I didn't try finding exact values.

Upd: Idk the reason for these downvotes, but it's correct. I have 100 points (and 6th rank) on the scoreboard using this solution and I'm confident about its correctness as well. See you all in Finals now.

Code

    vii dp(n+1,mp(-INF,-INF));
    dp[0].Y=0;
    for(lli i=1;i<=n;++i){
        auto &cur=dp[i];
        for(lli j=max(0LL,i-20);j<i;j++){
            lli sum=0;
            for(lli k=1;k<=i-j;++k){
                if(k&1)
                    sum+=a[j+k];
                else
                    sum-=a[j+k];
            }
            cur.X=max(cur.X,dp[j].Y+sum);
            cur.Y=max(cur.Y,dp[j].X-sum);
        }
    }
    cout<<max(dp[n].X,dp[n].Y)<<endl;

_manvirag

← Rev. 3 →

Is there any way , to get notify on new blog in recent actions in Codeforces.

:(, feeling very bad not able to participate , I did not know about contest .

mystery__coder

+17

Has anyone received any mail saying "Congratulations..etc and asking for details for sending the prize"?

It's been more than 3 days and not a single mail from gfg after the contest. Ashishgup any clue?

Swagat_nhi_karoge_hmara

Any updates on prize distribution Ashishgup,Omja?

akshaywairagade2

Will the contestants above rank 200 be included under 200 after removing overseas particpants.

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