Hint: you should only know fundamental theorem of arithmetic

Треугольное число

The question stated that 0 < N < 30.  However, in test case #9 N is exactly 30, contradicting the constraint.

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First I got WA#22 while I used double
than I change to use longdouble instead of double and I got WA#24
After that I change floor(x)/ceil(x) to floor(x+esp)/ceil(x-esp) and I got WA#24 too
But when I try to use double again I got AC!
[It's strange,too strange.I hate Accuracy issues]
(Sorry for my poor English XP)

Edited by author 07.09.2021 16:36

I guess that case is sth like a==0(a=-5Nz) ,b!=0 (b= VxNx+VyNy+VzNz)
(I find a check t=c/b instead of t=-b/c by mistake,and got wa)

there are 24 unique sequences of edges for the specific cube, i use 12 sequences and flip flop front with back for each of them
    static int R[][] = new int[][] {
        {1,2,3, 4,5, 6},
        {5,3,1, 4,2, 6},
        {2,1,5, 4,3, 6},
        {3,5,2, 4,1, 6},

        {6,4,3, 1,5, 2},
        {5,3,6, 1,4, 2},
        {4,6,5, 1,3, 2},
        {3,5,4, 1,6, 2},

        {1,2,6, 3,4, 5},
        {6,4,2, 3,1, 5},
        {2,1,4, 3,6, 5},
        {4,6,1, 3,2, 5}
    };
    static int front_back_rotate[][] = new int[][]{{0,1,2,3,4,5},{0,1,4,5,2,3}};
    //left right top  front bottom back
    //0       1     2    3     4      5
    static boolean eqv(int a[], int b[]) {
        for (int[] r: R) {
            for (int[] I: front_back_rotate) {
                int front_back[] = new int[6];
                for (int i=0; i<6; i++) {
                    front_back[r[I[i]]-1] = a[i];
                }
                if (Arrays.equals(front_back, b)) return true;
            }
        }
        return false;
    }

What is test 66?

Take attention:
No matter whose turn it is,if all the color is A,than Alise win,else Bob win.

Just like the title,I know how to solve this problem,becouse I have played the game before,and I know there is a way to move the chess.But if somebody dosen't play the game at all before,than how can he solve the problem?(Sorry for my poor English)

Edited by author 03.09.2021 16:00

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int main() {

   int n, k;

   scanf("%d %d", &n, &k);

   if (n < 1 || n > pow(10, 9) || k < 1 || k > pow(10, 9)) {
      return 0;
   }

   unsigned long long result = 0;
   unsigned long long buffer = 1;
   n--;

   for (int i = 0; n >= 0; i++) {

      n -= buffer;
      result++;
      buffer *= 2;

      if (n <= 0) {
         printf("%llu\n", result);
         return 0;
      }

      if (buffer >= k) {
         buffer = k;
         double count = (double)n / buffer;
         if (count != (long long)count) {
            count++;
         }
         result += count;
         printf("%llu\n", result);
         return 0;
      }
   }
   return 0;
}

Edited by author 21.01.2020 01:57

use hash :)

use dsu

Give me some tests.
Some my tests
Input:
3
9 2 1 3 1 4 1 1 1 5
3 1 2 1
3 1 2 1
Output:
G
1 2
Input:
1
5 2 3 1 4 5
Output:
G
1 3
Input:
1
5 5 4 1 3 2
Output:
G
1 1

Edited by author 25.09.2007 20:24

using System;
using System.Linq;
namespace e3
{
    class Program
    {
        public static void Main(string[] args)
        {
            int k=int.Parse(Console.ReadLine());
            int[] arr = Console.ReadLine().Split().Select(int.Parse).ToArray();
            int temp;
            for (int i = 0; i < k - 1; i++) {
        for (int j = 0; j < k - i - 1; j++) {
            if (arr[j] > arr[j + 1]) {
                // меняем элементы местами
                temp = arr[j];
                arr[j] = arr[j + 1];
                arr[j + 1] = temp;
            }
        }
    }

            int s=0;
            for (int i = 0; i < k/2+1; i++) {
                s=s+(arr[i]/2+1);
            }
            Console.WriteLine(s);

        }
    }
}

Try iterative instead of recursive. moves wont exceed 16.

Edited by author 25.08.2021 23:18

Edited by author 25.08.2021 23:19

I tried all test cases which were written in the discussion part and my program gives the correct answer for all of them. How to find my error:

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define endl "\n" /*DELETE IT ON INTERACTIVES!!*/
#define mod 1000000007
#define int long long
#define double long double
#define all(x) x.begin(), x.end()

const int N = 1e3+5;
vector<int> adj[N],marked(N,0),killed(N,0),vis_1(N,0),torun(N,1),vis_2(N,0);
int ok;

void dfs(int node){

    vis_1[node] = 1;
    if(killed[node]) ok = 1;

    for(auto go : adj[node]){
        if(!vis_1[go]){
            dfs(go);
        }
    }
}

void mark(int node){

    marked[node] = 1;
    vis_2[node] = 1;

    for(auto go : adj[node]){
        if(!vis_2[go]){
            mark(go);
        }
    }
}

void solve(){

    int n; cin>>n;

    string s,t;

    while(cin>>s){
        if(s == "BLOOD") break;
        cin>>t;
        adj[(int)s[0] - '0'].pb((int)t[0] - '0');
        torun[(int)t[0] - '0'] = 0;
    }

    int a;

    while(cin>>a){
        killed[a] = 1;
    }

    for(int i=1;i<=n;i++){
        if(torun[i]){
            dfs(i);
            if(ok){
                mark(i);
            }
            fill(all(vis_1),0);
            fill(all(vis_2),0);
            ok = 0;
        }
    }

    int hehe = 0;

    for(int i=1;i<=n;i++){
        if(!marked[i]){
            hehe = 1;
            cout<<i<<' ';
        }
    }

    if(!hehe) cout<<0;
}

int32_t main(){

    cin.tie(0); ios::sync_with_stdio(0);

    int t=1; //cin>>t;

    while(t--) solve();

    return 0;
}

5
1 2 3 4 5
shold be
4 1
1 2 3 4
5
(or others)
but can't be
5 0
1 2 3 4 5

What if one of them is outside the trap? Do they still need to keep contact (have at least one common corner)?