we dont know use the locomotive as a carriage, like:
input:
1 1
AA
AB

output:
NO

GOOD LUCK!!!

Whats wrong with anything here?
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using ui=unsigned int;
int main() {
    //ios::sync_with_stdio(0);
    //cin.tie(0);
    ll n;cin>>n;
    if(!n){
        cout<<10;
        return 0;
    }
    else if(n<10){
        cout<<n;
        return 0;
    }
    vector<int>v;
    ll l=n;
    for(bool q=1;q;){
        q=0;
        for(ll i=9;i>1;i--){
            if(l%i==0){
                l/=i;
                v.push_back(i);
                q=1;

            }
        }
    }
    v.push_back(l);
    sort(v.begin(),v.end());
    if(v[0]==1)for(int i=1;i<v.size();i++)cout<<v[i];
    else cout<<-1;
    return 0;
}

i had wa 22 for many times , & i finally found my fault & then got AC .
my fault is that i didn't find the exact edge (with the same ci & ti & hi) it used .
remember there are many different edges between a (ui , vi) .

let me know if there is a test case in that the any first number of A , B, C is 0

Please make clear that

Edited by author 16.12.2006 14:06

Edited by author 16.12.2006 14:06

Edited by author 16.12.2006 14:07

let N be the node num of the graph.
so why the maximum cycle number is M-N+1?
where M is the edge num.

please...

I found the problem statement a little hard to understand, and I want to make some clarifications for anyone who is having difficulties.

1. We are looking for the maximum number of cycles such that each of these chosen cycles has at least one edge which is not included in any other chosen cycle.

2. The graph is not necessarily connected.

I originally thought that we had to choose the maximum number of cycles which have at least one edge not included in all other chosen cycles. Hope this can help people.

Test #30 was found to be incorrect. It has been removed. The failing solutions have been rejudged.

Подскажите, пожалуйста, что не так с кодом? Wrong answer, тест 3.

Edited by author 09.10.2022 19:48

Edited by author 09.10.2022 19:48

Amount of pieces to move, edges to make euler path(only pieces with (in[i] + out[i]) > 0) and amount of components

Try this test (if you look neighbour cells in order from -1 left -1 up (11 oclock) clockwise)
4 4
#A#.
#A!.
##.#
.*$.

dfs or bfs is the correct idea?
oh, i'm done with it (:
Edited by author 15.10.2012 15:14

Edited by author 15.10.2012 15:22

Если я правильно понимаю, надо посчитать количество яиц, необходимых для поиска E бинаркой, а если оно окажется больше количества яиц в гнезде, то вывести количество этажей?

#include <iostream>
#include <vector>
#include <iomanip>
#include <cmath>
using namespace std;

int main (){
    double Input=0 ;
    double number[] ;
    int counter=0;

    while ( cin >> Input && Input != EOF){
        number[counter] = Input;
        cout << number[counter];
        counter++ ;
    }

    for (int i=counter ; i>0; i--){
        cout << setprecision(4) << sqrt (number[counter]) << endl;
    }
}

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

signed main(){
  int n;
  cin>>n;
  int sum=0;
  for(int i=0;i<n;i++){
    int x;
    cin>>x;
    sum+=x;
  }
  double cnt1=(double)sum/(double)n;
  if(cnt1==3){
    cout<<"None";
  }else if(cnt1==5){
    cout<<"Named";
  }else if(cnt1>=4.5 && cnt1!=5){
    cout<<"High";
  }else if(cnt1!=3 && cnt1!=5 && cnt1<4.5){
    cout<<"Common";
  }
}
getting a wrong answer on test 8

i use clear binmask as in 1122, but in 1122 u can chek for answer your board by every change of bit, BUT in THIS problem, u need check board when you finish your changes with bits.
Example u have bits 1101 and next is 1110,
u firstly change 1111 and then 1110 (two iterations because we need to do this iterations on board)
so if u do check on every that iteration its wa19

this is can be possible

On test
bwwb
bbwb
bwwb
bwww

My program gives 3, but right ans is 4
this test from 1060

//Input:
12 10
0 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 1 0 0 0
0 0 2 3 1 1 1 0 0 0
0 0 2 4 2 0 0 0 0 0
0 0 0 2 3 1 0 0 0 0
0 0 0 0 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0

//Correct output:
WWWWW
WRWWW
WWRWW
WWWWW
WWWWW
WWWWW

WRRW
WRWW
WWWW
WWWW
WWWW

RWW
WWW
WWW
WWW

RW
WW
WW

W
W