#include "bits/stdc++.h"

#define int long long
using namespace std;

vector<int> eratosfen() {
    int n = 168841;
    vector<bool> res(n + 1, false);
    for (int i = 2; i <= n; ++i) {
        if (!res[i]) {
            for (int j = i * i; j < n; j += i) {
                res[j] = true;
            }
        }
    }
    vector<int> res2;
    for (int i = 2; i < res.size(); ++i) {
        if (!res[i]) {
            res2.push_back(i);
        }
    }
    return res2;
}

signed main() {
    int t;
    cin >> t;
    vector<int> a = eratosfen();
    while (t--) {
        int n;
        cin >> n;
        if (n == 1) {
            cout << 2;
            continue;
        }
        cout << a[n - 1] << endl;
    }
}

Use cumulative count of sorted values. int64_t is not needed (e.g.WA3).

It's very strange but it seems that even long double precision doesn't enough for this problem. I had function get_x(polyline, y), that calculates x coordinate of polyline on coordinate y. I implemented it through binary searching and then calculating by formula, but that resulted in WA23. Then I made an optimisation: if polyline has integer point with coordinate y: (x, y), I instantly return x.

Edited by author 03.01.2024 20:12

I have done bs on overspeeding with upper bound of 1e12 and EPS of 1e-9.
Any idea?
Regards
So Sui Ming

https://en.wikipedia.org/wiki/Prime_number_theorem

Prime numbers meets often. You can check that head-on decision (check every a XOR i, i=0,1,2...) uses a few numbers

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;
    }

}