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?

#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;
}

Использую метод отжига и у меня ТЛ 16 тест, в кф проверил что у меня на макс тесте работает не более 2с. Возник вопрос, админы специально поставили такое ограничение чтобы метод отжига не проходил ?

#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <deque>
#include <queue>
#include <algorithm>
#include <iterator>
#include <bitset>
#include <iomanip>
#include <functional>
#include <list>
#include <cctype>
#include <array>
#include <algorithm>

using namespace std;
int main() {
    int n;
    n = 5;
    vector<vector<int>> vec(n);
    for (int i = 0; i < n; ++i) {
        vec[i].resize(n);
        for (int j = 0; j < n; j++)
        {
            cin >> vec[i][j];
        }
    }
    cout << min(min(min(vec[0][0] + vec[1][0] + vec[2][1] + vec[2][3] + vec[3][4],
                vec[0][0] + vec[3][0] + vec[2][3] + vec[1][2] + vec[1][4]),
                vec[0][0] + vec[0][2] + vec[1][2] + vec[1][3] + vec[3][4]),
                vec[0][0] + vec[0][2] + vec[2][3] + vec[1][3] + vec[1][4]) << endl;

    if (min(min(min(vec[0][0] + vec[1][0] + vec[2][1] + vec[2][3] + vec[3][4],
        vec[0][0] + vec[3][0] + vec[2][3] + vec[1][2] + vec[1][4]),
        vec[0][0] + vec[0][2] + vec[1][2] + vec[1][3] + vec[3][4]),
        vec[0][0] + vec[0][2] + vec[2][3] + vec[1][3] + vec[1][4]) ==
        vec[0][0] + vec[1][0] + vec[2][1] + vec[2][3] + vec[3][4])
    {
        cout << "1 2 3 4 5";
    }
    else if(min(min(min(vec[0][0] + vec[1][0] + vec[2][1] + vec[2][3] + vec[3][4],
        vec[0][0] + vec[3][0] + vec[2][3] + vec[1][2] + vec[1][4]),
        vec[0][0] + vec[0][2] + vec[1][2] + vec[1][3] + vec[3][4]),
        vec[0][0] + vec[0][2] + vec[2][3] + vec[1][3] + vec[1][4]) ==
        vec[0][0] + vec[3][0] + vec[2][3] + vec[1][2] + vec[1][4])
    {
        cout << "1 4 3 2 5";
    }
    else if (min(min(min(vec[0][0] + vec[1][0] + vec[2][1] + vec[2][3] + vec[3][4],
        vec[0][0] + vec[3][0] + vec[2][3] + vec[1][2] + vec[1][4]),
        vec[0][0] + vec[0][2] + vec[1][2] + vec[1][3] + vec[3][4]),
        vec[0][0] + vec[0][2] + vec[2][3] + vec[1][3] + vec[1][4]) ==
        vec[0][0] + vec[0][2] + vec[1][2] + vec[1][3] + vec[3][4])
    {
        cout << "1 3 2 4 5";
    }
    else {
        cout << "1 3 4 2 5";
    }
    cout << endl;
}

var n,k,ch,im:int64;
begin
read(n,k);
n:=n-1;
ch:=0;
im:=1;
while n>0 do begin
  if im<=k then begin
    n:=n-im;
    im:=im*2;
    ch:=ch+1;
    end;
  if im>k then begin
    n:=n-k;
    im:=im+k;
    ch:=ch+1;
    end;
  end;
writeln(ch);
end.

I hope these tests will be useful for you :)
4 5 0
3 4
1 2
3 1
2 4

and

7 7 0
1 2
1 3
2 4
2 5
3 6
3 7
4 7

The output must be 0 in both cases, but if you've done something wrong, it can be -1. Make sure your determination of connected components isn't wrong.

vector<vector<int64_t>> sequence = {
  {1,1,1,1,1,1,1,1,1,1,1,1},
  {1,0,2,0,4,0,8,0,16,0,32,0},
  {1,2,6,14,37,92,236,596,1517,3846,9770,24794}, // next is 62953
  {1,0,14,0,154,0,1696,0,18684,0,205832,0}, // next is 2267544
};

The solution is a linear recurrence. Output is multiplied by 2.

Edited by author 16.01.2020 16:32

I wrote a stupid solution and got wa32. Help please!

In my case this test may help you

16 19
1 2
2 3
3 4
4 5
5 1
1 6
6 7
7 8
8 9
9 10
10 6
5 11
11 12
12 13
13 4
3 14
14 15
15 16
16 2

The sequence of contraction affects the solution if you do not repeat the process after release the blossom.

if you contract (6,7,8,9,10), (4,5,11,12,13), (2,3,14,15,16) first
   you get maximum matching.
but if you contract (1,2,3,4,5) first and don't repeat the process after release it
   you don't get to the maximum.

Hope this helps.

Edited by author 13.01.2020 19:23

Edited by author 13.01.2020 19:23