Test case #6:

9 40
10 10 10 10 10 10 10 10 10 10
0 5 6 1 0 0 0 0 0 0
12 20

Answer: 158882255


Test case #12:

10 20
52 4 2 3 6 7 6 3 6 5 6
6 0 0 0 0 1 0 0 0 0 1
10 50


Answer: 978217740

PyCharm calculates n=2000 in 0.55 sec while here I got exceed time limit with 2.020 sec. I don't get it.

Can't find the error, because it gives the same answers as those stupid precalc solutions:

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

int soma[40] = {0};

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, i, lim;
  ll ans = 0;
  cin >> n;
  lim = pow(10, n/2);
  for(i=0; i<lim; ++i){
      soma[i%10 + (i/10)%10 + (i/100)%10 + (i/1000)%10]++;
  }
  for(i=0; i<=n*9; ++i){
      ans += soma[i]*soma[i];
  }
  cout << ans << '\n';
  return 0;
}

#include<bits/stdc++.h>
using namespace std;
int n,m,q,a[100003],b[100003],c[100003];
vector<pair<int,int> >v;
vector<int>g[100003];
bool vis[100003];
int qq[100003];
map<int,int>mp;
void dfs(int x,int cc){
    vis[x]=1;
    for(int i=0;i<g[x].size();i++)
        if(!vis[g[x][i]])
            dfs(g[x][i],cc);
    qq[x]=cc;
}
int main(){
    cin>>n>>m;
    for(int i=0;i<m;i++)
        cin>>a[i]>>b[i],
        a[i]--,
        b[i]--;
    cin>>q;
    for(int i=0;i<q;i++)cin>>c[i],c[i]--,v.push_back(make_pair(a[c[i]],b[c[i]])),a[c[i]]=b[c[i]]=-1;
    reverse(c,c+n);reverse(v.begin(),v.end());
    for(int i=0;i<m;i++)
        if(a[i]!=-1)
            g[a[i]].push_back(b[i]),
            g[b[i]].push_back(a[i]);
    int cnt=0;
    for(int i=0;i<n;i++)
        if(!vis[i])
            mp[cnt]=cnt,
            dfs(i,cnt++);
    vector<int>ans;
    for(int i=0;i<q;i++){
        ans.push_back(cnt);
        if(mp[qq[v[i].first]]!=mp[qq[v[i].second]])
            mp[qq[v[i].first]]=mp[qq[v[i].second]],
            qq[v[i].first]=qq[v[i].second],
            cnt--;
    }
    reverse(ans.begin(),ans.end());
    for(int i=0;i<ans.size();i++)
        cout<<ans[i]<<' ';
}

JavaEntryPointWrapper.java:10: error: cannot access Oppa
            Oppa.main(args);
            ^
  bad source file: .\Oppa.java
    file does not contain class Oppa
    Please remove or make sure it appears in the correct subdirectory of the sourcepath.
1 error
Выходит такая ошибка, почему?

Please give this test!
lifanov@mail.ru

Please, delete my account. I don't need it anymore.

Please, delete my account. I don't need it anymore.

but I couldn't find any way to delete the post

Edited by author 27.01.2020 01:24

Edited by author 27.01.2020 01:24

Edited by author 27.01.2020 01:24

I've passed all the tests the forum had. Can someone give me more

Try the following 2 tests:

2
0
10000
2
20000
10000

and

2
0
10000
2
0
-10000

What answer is issued by your program?

Edited by author 14.02.2013 20:37

Very easy:)

*first if k>n/2 make it smaller n/2: k=n-k+1

*then the answer is (n-k-2)

 (......[Vasya].......[n-1][n]
  if [n-1] and [n] sit,then all people between
  [Vasya] and [n-1] will stumble at Vasya.
  And this will be the maximal answer )

*special case: n=2 => answer is 0

n = 1

Hello

My solution failed on test 21. Passed next cases:

5 5
2 2 2 3 4
0 0

5 4
4 1 4 1 2
3 2

5 4
2 2 1 1 1
5 2

5 7
4 1 4 1 2
3 1

10 18
2 2 2 2 2 2 2 2 2 2
0 0

If you have any idea please let me know. Thank you

Edited by author 21.08.2015 10:49

#include <iostream>

using namespace std;

int main()
{
        int n, k;
        int i;
        int show, type;
        int showTotal = 0, typeTotal = 0;
        int neededType;

        cin >> n >> k;
        for ( i = 1; i <= n; i++)
        {
                cin >> show >> type;

                if (show - type == 2)
                        showTotal = typeTotal = 0;
                else
                {
                        showTotal += show;
                        typeTotal += type + 2;
                }
        }
        neededType = k + showTotal - typeTotal - 2;
        if (neededType > 0 && neededType < 101)
                cout << neededType << endl;
        else
        {
                cout << "Big Bang!" << endl;
        }
        return 0;
}

#include <iostream>
using namespace std;
int main()
{
    int n,a[100],b[100],c,d,i,j;
    cin>>n;
    for(i=0;i<n;i++)
        cin>>a[i]>>b[i];
    for(i=0;i<n;i++)
        for(j=i+1;j<n;j++)
            if(b[i]<b[j])
            {
                c=b[i];
                b[i]=b[j];
                b[j]=c;
                d=a[i];
                a[i]=a[j];
                a[j]=d;
            }
    for(i=0;i<n;i++)
        cout<<a[i]<<' '<<b[i]<<'\n';
    return 0;
}


Help me! I'm die.
Sorrry,my English is poor.
I will try my best to understand you.

Try this simple test:
11123
Answer:
111

Please help with wrong answer 7.
Possible input?

my solution is the following:
start with least common multiple = 1
make char array of 1000 000 to track forbidden divisors

let div be the divisor and ans the answer

if ans == 1 make new lcm of the previous lcm and current divisor. if it is over 10^12 - fail. find all the divisors of div, check that they are not forbidden in the forbidden array
if ans == 0 - check if lcm is divisible by the divisor and fail if it is.

complexitiy O(n * sqrt(1000000)) (sqrt from the "find all divisors)

my solution gets 0.5 sec (using cin,cout with tie(null))
what is the solution that gets less than 0.1 sec? Is there any hint?

#include<stdio.h>
#include<math.h>
int main()
{
    unsigned long long int a[1000],i=0,b,c;
    double s;
    while(scanf("%llu",&b)!=EOF)
    {
        a[i]=b;
        i++;
        c=i;
    }
    for(i=c;i>=0;i--)
    {
        s=sqrt(a[i]);
        printf("%.4lf\n",s);
    }
    return 0;
}