Common Board @ Timus Online Judge

coder To ADMIN add test, because tests are weak [1] // Problem 2175. Staircase 25 May 2025 13:20

🦄imosk72🦄[GTGU] Strange statement [1] // Problem 2181. Student, or There and Back Again 9 Mar 2025 16:50

bidzilya Test 7 contains intersecting squares which contradicts the statement [1] // Problem 1097. Square Country 2 10 May 2025 20:01

Juve Input Value Exception [1] // Problem 1644. A Whole Lot of Walnuts 12 Apr 2025 12:25

adhamet Problem Editorial (written by a beginner) [1] // Problem 2025. Line Fighting 18 Oct 2025 15:00

First of all, we need to understand which number of teams x (from 2 to k) we need to pick to distribute our fighters into.

Through intuition we might get the idea that the bigger the K **the more the fights** (which is what is required).
It can be proven that our intuition is correct; the more you divide n the better the result, therefore we will need to divide n by k.
Proof: assuming we have n = 3, now let us assume we will distribute them to 1 team where the first team is of 3 fighters, there will be 0 fights, let us now assume we will distribute them to 2 teams where the first is of 2 and second is of 1 fighters, there will be 2 fights, lastly let us assume that we will distribute them to 3 teams where each team has a fighter, there will be 3 fights, we can then deduce that for each fighter you divide into a new group more fights are formed.
Conclusion: It is always the best choice to divide by the biggest possible number you can divide by, thus we will always divide by K.

To solve the problem we can then deduce that there will always be two cases:
Case 1 (k divides n): here we simply divide fighters into k groups, each group (composed of (n/k) fighters) will fight some other group (composed of (n/k) fighters) thus the equation for the first case is simply:
(k * (k - 1) / 2) * (n / k)^2

Case 2 (k doesn't divide n): here we notice that n will be composed of x count of floor(n / k) and y count of ceil(n / k) meaning floor(n/k)*x + ceil(n/k)*y = n, and that obviously, x + y = k, through these equations we can deduce algebraically the values of x and y, and from it form our equation.
Our equation will have x groups of floor (n / k) fighters which within them will form a specific number of fights, we will also have y groups of ceil (n / k) fighters which within them will form a specific number of fights, then we will also need an to find an equation that calculates the number of fights that happens between the x groups and the y groups, through observation and deduction you will find that this will be the equation of the second case:
((x * (x - 1) / 2) * (floor(n/k))^2) +
((y * (y - 1) / 2) * (ceil(n/k))^2) +
(floor(n/k) * ceil(n/k)) * x * y

Code:
```
#include <bits/stdc++.h>

using i64 = long long;

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int t;    std::cin >> t;
    while (t--) {
        int n, k;    std::cin >> n >> k;
        if (n % k == 0) {
            int count = k * (k - 1) / 2;
            std::cout << count * (n / k) * (n / k) << '\n';
        } else {
            int x = n - ((n + k - 1) / k) * k;
            x = x / ((n / k) - ((n + k - 1) / k));
            int y = k - x;

            int part1 = (x * (x - 1) / 2) * (n / k) * (n / k);
            int part2 = (y * (y - 1) / 2) * ((n + k - 1) / k) * ((n + k - 1) / k);
            int part3 = (n / k) * ((n + k - 1) / k) * x * y;
            std::cout << part1 + part2 + part3 << '\n';
        }
    }

    return 0;
}
```

Edited by author 18.10.2025 15:04

Edited by author 18.10.2025 15:06

Edited by author 30.10.2025 12:18

adhamet Re: Problem Editorial (written by a beginner) // Problem 2025. Line Fighting 18 Oct 2025 15:03

Thank you for reading my editorial.

Edited by author 18.10.2025 15:05

~'Yamca`~ WA 42 // Problem 1043. Cover an Arc 16 Oct 2025 21:12

~'Yamca`~ WA 2 // Problem 1043. Cover an Arc 16 Oct 2025 21:11

gaoshangbo test 2 ？？？？？！！！！！！ [2] // Problem 1168. Radio Stations 4 Oct 2011 19:43

RAVEman Please help [1] // Problem 1168. Radio Stations 1 Jul 2007 04:17

Where am i wrong? It should be obvious :) because it is WA2

#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <iostream>
#include <fstream>
#include <iterator>
#include <math.h>
#include <cstdio>
#include <cstdlib>
#include <sstream>

using namespace std;
const double eps=1e-6;

struct R{
    int x;
    int y;
    double r;
};

R r[1111];
int a[55][55];
bool s[55][55];
int n,m,k,x,y;

int main(){
    cin>>n>>m>>k;
    for(int i=0;i<m;i++)
        for(int j=0;j<n;j++)
            cin>>a[i][j];
    for(int i=0;i<k;i++){
        cin>>r[i].x>>r[i].y>>r[i].r;
        r[i].x--;r[i].y--;
        s[r[i].x][r[i].y]=true;
    }
    long res=0;
    for(int i=0;i<m;i++)
        for(int j=0;j<n;j++)
            if(!s[i][j]){
                bool possible=true;
                double d;
                int mina=a[i][j];
                int maxa=32000;
                int _mna,_mxa;
                for(int q=0;q<k;q++){
                    d=r[q].r*r[q].r-(i-r[q].x)*(i-r[q].x)-(j-r[q].y)*(j-r[q].y);
                    if(d<-eps) {possible=false;break;}
                    d=floor(sqrt(d+eps))+eps;
                    _mna=d;_mna=a[r[q].x][r[q].y]-_mna;
                    _mxa=d;_mxa=a[r[q].x][r[q].y]+_mxa;
                    if(mina<_mna) mina=_mna;
                    if(maxa>_mxa) maxa=_mxa;
                    if(mina>maxa) {possible=false;break;}
                }
                if(possible)
                    res+=(maxa-mina+1);
            }

    cout<<res<<endl;
    return 0;
}

Edited by author 01.07.2007 04:19

ViktorGostev91 Re: Please help // Problem 1168. Radio Stations 16 Oct 2025 20:48

replace cin>>n>>m>>k; with cin>>m>>n>>k;

coder My O(n) loop solution AC 0.001s, but O(1) sum solution AC 0.015s :)) // Problem 1068. Sum 16 Oct 2025 09:07

Vladimir Yakovlev (USU) Problem 1281. River Basin reworked // Problem 1281. River Basin 16 Oct 2025 03:44

Keworker `~ If you have WA#47 [1] // Problem 2102. Michael and Cryptography 14 Jul 2024 11:03

Imback the last digit of N...... // Problem 1049. Brave Balloonists 11 Oct 2025 12:45

Imback use long long but not double // Problem 1011. Conductors 10 Oct 2025 13:10

~'Yamca`~ If you have WA6 // Problem 2163. Interview 23 Sep 2025 21:38

Bob WA2 Помогите! // Problem 1226. esreveR redrO 17 Sep 2025 00:00

zser Brute force Accept in 1.625 [1] // Problem 1044. Lucky Tickets. Easy! 8 Jun 2025 20:41

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

int cal(int x){
    int ret = 0;
    while (x > 0){
        ret += x % 10;
        x /= 10;
    }
    return ret;
}

bool lucky(int x, int mm){
    int a = x / mm;
    int b = x % mm;
    if (cal(a) == cal(b)) return true;
    return false;
}

int main(){
    int n;
    cin >> n;
    int nn = round(pow(10, n)) - 1;
    int mm = round(pow(10, n / 2));
    int tot = 0;
    for (int i = 0; i <= nn; i++){
        if (lucky(i, mm)) tot++;
    }
    cout << tot << endl;
}

if your code is simple and fast enough, brute force can work.

codeJutsu Re: Brute force Accept in 1.625 // Problem 1044. Lucky Tickets. Easy! 14 Sep 2025 22:58

the input N consists of only even numbers, meaning the number of digits won't be odd.
moreover, input range is limited between 2 and 8, meaning the only possible inputs are 2, 4, 6 and 8.

now this would be my approach :

lets say, the input number is 4 digits in length. meaning, the sum of the first 2 digits need to be equal to the sum of the last 2 digits.

what i'd do is, i'd check for all 2-digit numbers starting from 00 all the way to 99, and store the sum of their digits in an array (in this particular case there are total 100 numbers so that'd be the array length).

after storing the sums in an array, i'd run nested loops through the array. the outer loop would take a sum value from an index. the inner loop would search through all elements in the array from the beginning to the end (including that index from the outer loop too) to see if any sum value matches the outer loop's index sum value. if there's a match, the count increases by 1. the final value of the count is the answer. this process of nested loops is time consuming, so we can sort the array first and then only run the inner loop as long as the next sum values match with the current sum value. this will only improve the performance slightly, as there will still be cases that need the inner loop to run for longer.

a far better approach would be to find the maximum possible value of the sums, and then take an array of that size plus 1. for our example of 4-digit numbers, the maximum possible sum of 2-digit numbers would be 9+9=18 (or 9x2=18) because 99 is the largest 2-digit number. thus the array size here would be 18+1=19 (so that we can access from array[0] all the way to array[18] when needed).

initially, all index values of this array would be 0. as we go through all the 2-digit numbers, we'll find the sum of their digits and raise the value of the corresponding index in the array by 1. this way, we'll immediately know how many times a particular sum has occurred by just looking at that sum's index value, instead of running nested loops. for example, if array[4]=5, then the sum 4 has occurred 5 times in all 2-digit numbers (04, 13, 22, 31, 40). again, if array[5]=6, then the sum 5 has occurred 6 times (05, 14, 23, 32, 41, 50). all we need to do then is find the total sum of all the individual sum count's squares from the entire array.

in case someone doesn't understand why we need to square the individual sum counts : we only counted the sums for 2-digits in our example. but all the possible cases of the 2-digits for a particular sum will occur in both the left half and the right half of a 4-digit number. lets say, the sum value was 4 and a total of 5 cases were found where the 2-digits sum up to 4. then, for each of these 5 cases in one side, all of these 5 cases will occur too in the other side. thus, there will be 5x5=25 total cases where a 4-digit number will have the sum of its left half digits as 4 and the sum of its right half digits as 4 as well. if the sum value was 5, then we'd find 6x6=36 total such 4-digit numbers. that's why we square the individual sum counts before adding them up to the total sum.

there might be even better approaches that i'm not familiar with just yet. maybe an alien would be able to help you with such superior alien concepts.

nushrat Help me with the algorithm [1] // Problem 1044. Lucky Tickets. Easy! 19 Nov 2016 07:11

Hi,

I need desperate help with building the algo. It seemed easy at first, but I am getting all confused. Can anyone explain the step for calculation ?

Suppose the input is 4 digits, so that means I have to fill up 4 spaces with 9 digits(0-9) for each space and from the combinations, find the one that has equal sum for first 2 and last 2 digits.

If someone can help me with the steps, I think I can figure out the rest.

Thanks in advance.

codeJutsu Re: Help me with the algorithm // Problem 1044. Lucky Tickets. Easy! 14 Sep 2025 22:23

the input N consists of only even numbers, meaning the number of digits won't be odd.
moreover, input range is limited between 2 and 8, meaning the only possible inputs are 2, 4, 6 and 8.

now this would be my approach :

lets say, the input number is 4 digits in length. meaning, the sum of the first 2 digits need to be equal to the sum of the last 2 digits.

what i'd do is, i'd check for all 2-digit numbers starting from 00 all the way to 99, and store the sum of their digits in an array (in this particular case there are total 100 numbers so that'd be the array length).

after storing the sums in an array, i'd run nested loops through the array. the outer loop would take a sum value from an index. the inner loop would search through all elements in the array from the beginning to the end (including that index from the outer loop too) to see if any sum value matches the outer loop's index sum value. if there's a match, the count increases by 1. the final value of the count is the answer. this process of nested loops is time consuming, so we can sort the array first and then only run the inner loop as long as the next sum values match with the current sum value. this will only improve the performance slightly, as there will still be cases that need the inner loop to run for longer.

a far better approach would be to find the maximum possible value of the sums, and then take an array of that size plus 1. for our example of 4-digit numbers, the maximum possible sum of 2-digit numbers would be 9+9=18 (or 9x2=18) because 99 is the largest 2-digit number. thus the array size here would be 18+1=19 (so that we can access from array[0] all the way to array[18] when needed).

initially, all index values of this array would be 0. as we go through all the 2-digit numbers, we'll find the sum of their digits and raise the value of the corresponding index in the array by 1. this way, we'll immediately know how many times a particular sum has occurred by just looking at that sum's index value, instead of running nested loops. for example, if array[4]=5, then the sum 4 has occurred 5 times in all 2-digit numbers (04, 13, 22, 31, 40). again, if array[5]=6, then the sum 5 has occurred 6 times (05, 14, 23, 32, 41, 50). all we need to do then is find the total sum of all the individual sum count's squares from the entire array.

in case someone doesn't understand why we need to square the individual sum counts : we only counted the sums for 2-digits in our example. but all the possible cases of the 2-digits for a particular sum will occur in both the left half and the right half of a 4-digit number. lets say, the sum value was 4 and a total of 5 cases were found where the 2-digits sum up to 4. then, for each of these 5 cases in one side, all of these 5 cases will occur too in the other side. thus, there will be 5x5=25 total cases where a 4-digit number will have the sum of its left half digits as 4 and the sum of its right half digits as 4 as well. if the sum value was 5, then we'd find 6x6=36 total such 4-digit numbers. that's why we square the individual sum counts before adding them up to the total sum.

there might be even better approaches that i'm not familiar with just yet. maybe an alien would be able to help you with such superior alien concepts.

Edited by author 14.09.2025 22:42

IlushaMax Your computer won't explode if you will do like here))) [1] // Problem 1044. Lucky Tickets. Easy! 1 Apr 2016 23:55

SorrowAngel I HAVE THE CODE [2] // Problem 1044. Lucky Tickets. Easy! 23 Apr 2008 13:13