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

why is the 10th element 4? isn't it 3?

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.

I investigated on internet, but most of the answers were putting a type of character that breaks the cycle, but in the input example there´s no character and it´s not enough with a "while(cin >> n)" pls help

Try this simple test:
11123
Answer:
111

#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

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;
}

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 ?

My code is here:
--------------
#include <iostream>
using namespace std;

int compare(const void * x1, const void * x2) {
    return (*(int*)x1 - *(int*)x2);
}

int main() {
    long N;
    cin >> N;
    int *m = new int [N];

    for (int i = 0; i < N; ++i) {
        cin >> m[i];
    }
    qsort(m, N, sizeof(int), compare);

    if (N % 2 == 0) {
        cout << (((m[(N - 1)/ 2] + m[N / 2]) / 2.0)*10)/10;
    }
    else {
        cout << m[N/2];
    }

    return 0;
}
--------------
Tell me please where I made a mistake

#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;
}

Ya popitalsya perebrat' vse varianti dlya N = 6 i smog naschitat' tolko 14 vot oni:
1. WRWRWR    9. WRWRBW
2. RWRWRW   10. RWRWBR
3. WBRWRW   11. WBRBWR
4. RBWRWR   12. RBWBRW
5. WRBWRW   13. WRBWBR
6. RWBRWR   14. RWBRBW
7. WRWBRW   15. ??????
8. RWRBWR   16. ??????
Kakie dva ya upustil?

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.