Invitation to Biteration #2 (Rated for Div. 2)

I guess they will remove this feature and it is in their pipeline :slight_smile:

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@anon55659401 We’d also be having a varied score distribution for max points per problem.

Hey, please explain in detail, did not really understand that sentence.

Also, as @nagpaljatin141 is the organizer, contesr is gonna be awesome. We aren’t worried about anything :slight_smile:

@anon55659401 It means that points alloted for passing each test case for any problem may or may not be 100 pts. The Max points per problem (points after passing all test cases) would follow a varied distribution.

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So, it means every problem has a different value…just like cf right ?

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Technically, It won’t be CF, since points per problem won’t decrease with time, and there would be subtasks as well, but rest is same.

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But I hope all points for a problem will be mentioned like:-
1st problem-20;
2nd problem-40; etc…etc.

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By when will we get the tshirts?

I can’t guarantee any exact date, but we’d try to dispatch it to the winners as soon as possible.
For now, Enlighten us with any feedback for our future iterations of Biterations to come

Have you also participated?

@nagpaljatin141 look into this Algorithm to list solved problems in Partially Solved/ Fully Solved is incorrect

Please do something about server load. All the time, we were getting it and it reduces the rank for no reason.

Also, in the 3rd problem(MATHL), brute force gives TLE for 1st substask.
And, many things in the 2nd problem were unclear.
Example:–>
For this case:-
1 1 1 1 1
x=10…
Would ‘A’ eat up all the boxes or only “4” ?
Those who assumed 1 thing got AC and those who assumed other, got WA. It was more of dependent on luck.

According to my knowledge, everything in the 2nd Problem was explained in atleast one of the lines in the problem statement.
U can ask me any case, and I’d explain it from the problem statement itself.
About the case 1 1 1 1 1 1, A can’t eat all boxes, because both won’t reach the last box at the same time, B reaches at exactly 0 time, how can u assume that A would also reach there at 0 time, since he needs to eat 4 boxes involving 4 candies before?

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“A” can reach there in no time because he takes no time to move from 1 place to another!!!

(I asked you in comments, you replied to everyone except me!! :stuck_out_tongue: )

But it was nice contest :slight_smile:

Waiting for Biteration #3 :slight_smile:

But A starts eating candies from the leftmost box, and in the sample explanation also, he eats from left to right, so he can’t eat all the left candies in 0 time.

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Yes, I just wanted to know this in the contest. Thankyou!

Yup, for the same reason, we provided it in the problem statement itself

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Problem : B - Candy Love (BIT2B)
Can anyone let me know for which test case is my code failing. After the contest I did try to match my codes’s output to someone’s accepted solution on many random test cases and they are exactly matching. Yet my code is not getting accepted. Thanks in advance. :slightly_smiling_face:

My solution : solution

try this case

1
287
195 200 754 596 867 187 962 224 992 869 901 56 834 315 244 923 864 236 508 11 996 214 878 939 152 536 107 111 889 746 657 88 851 252 954 37 565 530 381 785 782 436 619 97 31 541 312 267 400 674 262 965 551 553 116 438 771 575 548 659 320 557 98 522 808 52 559 725 933 291 509 66 726 479 514 108 19 177 374 770 850 988 734 892 189 662 424 737 673 95 293 616 452 173 526 528 457 816 523 893 515 36 1 886 213 655 414 65 389 814 956 590 2 970 13 90 642 763 74 538 288 600 744 767 453 618 660 968 6 205 570 859 983 802 599 148 757 188 502 552 943 719 10 677 84 566 121 35 630 216 801 83 460 402 839 471 606 408 208 390 361 415 890 117 391 506 333 265 598 926 984 556 786 990 743 809 419 348 978 957 338 281 198 806 235 588 1000 92 921 617 633 714 636 539 466 421 675 455 271 549 199 689 896 766 18 762 355 43 272 160 629 211 512 720 483 128 325 193 232 828 123 139 797 945 342 67 493 740 9 124 853 435 830 895 706 875 268 594 751 332 866 524 854 891 704 976 773 349 191 169 240 650 260 249 112 683 955 7 296 233 827 109 831 449 974 302 761 278 416 920 764 592 440 241 699 366 811 989 80 615 401 439 114 227 547 944 27
1

output should be 142 145

Thanks again @syntaxhacker