My program stucked at WA-25 for several times and I don't know the reason why. One thing that confuses me is that I don't know how to deal with the following sentences in problem statement:
"You should consider a rotation to map a key point A to a key point B if A is mapped to a point within 10−6 units distance from the point B. Rotations having the same mapping should be considered as the same rotation."
I think the second sentence is redundant since a rotation is uniquely determined by its images on two different non diametral points. I don't know how to implement a function such that, given the images q_i of each point p_i, determine if any rotation f exists such that dist(f(p_i), q_i)<1e-6.

I had TL 5 because I did not add vertices to the BFS queue properly (sometimes I put the same vertex multiple times and then process in many times). You may try this test:

[test cut, too big]

Edited by moderator 06.09.2022 03:04

Firstly, I pre-calculate the index which carries 1, and store them on the map. then I take input and check the map if the map stored 1 for this input, then I printed 1, otherwise 0.
what is wrong here?

I walked through all topics in discussion, grabbed every test, but still got WA7.

I had several mistakes in my solution.

First of all, in some versions I forgot to make fixed format of output and forgot to cast my answer to integer type, therefore the solution printed the answer in scientific notation.

Test:
25
7217 6651
7358 2764
7034 5638
2431 6831
6550 2019
2334 1524
3375 4802
5960 4579
7078 1798
3176 7335
5872 3950
179 7351
9800 3982
2080 897
9345 2389
6470 6556
8420 9316
1671 7557
922 2613
9464 6735
1865 8925
6701 3365
6519 6665
9458 1535
8792 4383
Answer:
1495422

In some solutions there were rounding issues, I rounded down while should have rounded up.
Still I disagree that sometimes one should use floor(ans) instead of int(round(ans)), the latter always works fine in this problem.

Test:
3
4826 9656
137 7424
8340 2686
Answer:
22472

In some attempts I accidentally used = instead of ==, this test detected this perfectly.

Test:
4
0 0
2 0
2 2
0 2
Answer:
14


Edited by author 01.09.2022 22:29

ITS REALLY HARD
I WORK SO HARD FOR THIS PROBLEM TO DO ITS REALLY FAST, U HAVE 1MB MEMORY input THIS IS ABOUT 600-700k symbols, U NEED TO DO ITS Very VERY VERY VERYVERY VERy fAST AND THIS VERYV VEryV eVEYR Very HARD
U NEED TO READ THIS NEVER NEVER NEVER USE CIN
USE c=getchar() where c - char. And IF U WANT TO STOP INPUT JUST int WHILE
in the conditions of the while loop, put restrictions on the symbol, that is, there are all charms from A to Z and something else. AND THEN YOU NEED A REALLY FAST ALGORITHM
I rewrote mine six times, I can't understand it myself, it's very difficult, I only remember that char is int, and int is a deadline, and diffirintly THE ALGORITHM is SLOW

Please remember that input can contain numbers with more than one digit. My WA was due to this, and I don't think anybody will be this silly. But still posting! :)

Edited by author 29.08.2022 21:23

Edited by author 29.08.2022 21:23

delete iostream and using namespace std.
all cin cout replace on scanf and pritf
write in start of code: #define CRT_SECURE_NO_WARNINGS
and #include<cstdio>
and behind your vectors,pair and queue write std::
do vector WITHOUT ANY INTS, ALL BOOL
DONT USE PUSH_BACK its really slow, just write size of vector
Example:
>std::vector<std::vector<bool>> a(n);
>std::vector<bool> b(m);
and call with a[i][j] or something else;
If you look at other aspects of the task, then EASY BFS

String can have size 1000, max size is not 999.

I don't know, what is the maximum possible sum of digits, but in these tests it is 31.
Simple, badly optimized, recursion passes quickly.

Как узнать, когда игрок сделал страйк или спэр, если количество бросков во фрейме не задано?

I'm tried to solve it many times and this problem dont need strong. There is no need to count a lot, compare numbers, check so that all the money can be divided, all you need is to take all the powers of three and check for each that in sum with n it also gives the sum of the powers of three. That is, t is the sum of degrees, if t+n too, then output the answer immediately!

Be careful, the input numbers are float, though sample and all tests here have only integer numbers. It appears first at <= 6th test.

wa4 have test n=1 and some m
if u use dp and u write in your massive some int, when u write, u delete int and create new ints. But if u have n=1, u now answer for every for n=1 and when u use dp its write 0, it enters zero instead of the answer you wrote down, although there should be one in some cells

Just curious did anyone managed to pass the time limit
using binary search + rolling hash or the test data are
specially designed to be solved in time only by an efficient suffix tree/array implementation?

please answer to me.

#include<iostream>
#include<cmath>
#include<cstring>
long long int n;
int main(){
    const int k = 256;
    std::string g = "GOOD";
    std::string b = "BAD";
    std::string str;
    std::cin>>str;
    std::cin>>n;
    if( n == 0){
        std::cout<<"0"<<"\n";
        return 0;
    }
    int arr2[4];
    int bit = 1;
    int arr[4];
    long long int l = 0;
    memset(arr,0,sizeof(arr));
    memset(arr2,0,sizeof(arr2));
    for(int i = 3;i >=0 ; i--){
        long long int j = pow(k,i);
        for(int C = 1;C<=255;C++){
            long long int h = C*j;
            if( n >= h){
                l = n % h;
                if(l==0 && i!=0){
                    arr[i] = 1;
                    bit = 1;
                    n = n-bit*h;
                    break;
                }
                else if( i == 0 && n <255){
                    arr[i] = n;
                    break;
                }
                else {
                    arr[i] = n/h;
                    bit = n/h;
                    n = n-bit*h;
                    break;
                }

            }
            else break;
        }
    }

    if(str == g){
        long long sum = 0;
        for(int i = 0,m = 3; i<=3,m>=0;i++,m--){
            sum+=pow(k,m)*arr[i];
        }
        std::cout<<sum<<"\n";
    }
    else if( str == b){
        long long int sum = 0;
        for(int i = 0,m = 3; i<=3,m >= 0;i++,m--)
            sum+=pow(k,m)*arr[i];
        std::cout<<sum<<"\n";
        }
    else;
    return 0;
}

if u have wa 44 just remove one in the cyclic factorization, if there is one.
<=sqrt() + 1   -- bad
<=sqrt()       -- good

Can anybody help?

This was overall a great problems set!

Can anyone give a hint on the solution for "J. Turning Turtles"?
What special property (beside that it's a tree) must one use?
What is the complexity of the intended solution?

Thanks in advance!