Can you send to me some test????????))

my idea:

 sum( (-1)^m *  a^(n/(p[k1]*p[k2]*..*p[km] ) ) ), where p[i] - is prime number and n % p[i] == 0.

Domain and name must consist only of LOWERCASE letters and dots

The length of each crystal's side is L, not one.

4 строки 274 байта
кто короче?

Unfortunately, I can't write the Python solution due to the rules, but it exists! Try to find it!

input:
-2 -2 2 -2 2 2 -2 2
0 2 1
0 -3
output:
7.236067977

input:
1 0 0 1 -1 0 0 -1
0 0 1
2 -1
output:
1.236067977

 ... airplane must climb to an altitude of h meters during the first t seconds of the flight ...
Can time be less than t when height h is attained ?

Edited by author 18.12.2023 11:15

Edited by author 18.12.2023 11:15

1
10000
2
1 6

              1    2    3
1->2:180     180 -180
1->3:180     180      -180
2->3:180          180 -180
overall:     360   0  -360
prefix sum:  360  360   0

hence max from 1 to 3 is 360
do this similarly from 3 to 1

f(x) = (x-7)^2 + (x-4)^2 + (x-5)^2
Find x such that f(x) (cost function) is a minimum.

df/dx = 2(x-7) + 2(x-4) + 2(x-5) = 0
3x = 7+4+5
x = (7+4+5)/3

The biggest hurdle is the problem statement itself!

i think my algo is right but i have problem with overflow in C++.
who can give me good test?

PS: on Pypy i have TL50 using the same algo( And it's very strange because on test
1000000000 18
1 1
2 1
3 1
4 1
5 1
6 1
7 1
8 1
9 1
10 1
11 1
12 1
13 1
14 1
15 1
16 1
17 1
18 1

my algo works <300ms




PS: AC in C++ after many hacking with overflow. I hate this problem because spent a lot of time to fix problem with arighmetic (not algo)

Edited by author 12.12.2023 00:41

It seems that the worm always starts at back cover and ends at front cover
and order of volumes matters.

#include <iostream>

using namespace std;

int main()
{

    double k;
    while(cin >> k) {
        cout << fixed << setprecision(4) << sqrt(k) << endl;
    }

}

Подскажите, что нужно сделать чтобы у меня показывало на каком я месте в рейтинге?

Просто интересно, уже год-два моя страничка вне рейтинга

this is a not correct setting task, there are many solutions some of them is AC some of WA, but all of them are right

1 10 1 2 -> 2: means books are placed upside down, in this case first sheet of the first book is adjacent to the second book
1 10 2 1 -> 22: means books are placed correctly,  last sheet of the first book is adjacent to the second book

Edited by author 10.11.2015 07:42

Try this test
1 100
4
1 1
2 1
3 1
4 1
1 2
3 4
0 0
0 1
5 1

Correct answer is:
3.0200000
4 1 2 3 4

#include<bits/stdc++.h>
using namespace std;
bool isPrime(int n)
{
    if(n!=2&&n%2==0)return false;
    if(n<2)return false;
    for(int i=3;i<=sqrt(n);i+=2)
    {
        if(n%i==0)return false;
    }
    return true;
}
int main()
{
    int n;
    cin>>n;
    int x=9,r;
    if(n==1)cout<<1<<endl;
    else if(n==0)cout<<10<<endl;
else{
    vector<int>v;
    while(1)
    {
        if(n%x==0)
        {
           if(isPrime(n)==true)
           {
            if(n>9){cout<<-1<<endl;return 0;}
           }
           n=n/x;
           if(x==1)break;
           v.push_back(x);
        }
        else
        {
            x--;
        }

    }
    for(int i=v.size()-1;i>=0;i--)
    {
        cout<<v[i];
    }
    cout<<endl;
}
}

For large N, 2% of sample are counted.