a = int(input())
b = int(input())
if a %2==1 and b%2==1:
    print(int((b-a)/2+1))
else:
    print((b-a)//2)


Why WA??

Why does

bool comp(pair<int,int>& a, pair<int,int>& b)
{
    return a.second > b.second;
}

not AC with common sort?

How can the program recognize when there aren´t more inputs if use python3 ?
my program:
nums = [int(x) for x in input().split()]
nums.reverse()
for num in nums:
    print('{:.4f}'.format(num ** 0.5))

What's wrong? anyone can tell me ,pls

Edited by author 29.11.2023 07:42

Edited by author 29.11.2023 07:43

Consider 2 cases:
(1) K >= N
(2) K < N

and fry one side of each steak first.
Try cases like:
N=7,K=1; N=7,K=2; up to N=7,K=6 and find the pattern.

50 50 10
qwertyuiop[]\asdfghjkl;'zxcvbnm,./ QWERTYUIOP{}|AS
aegaegu
etoijqt
tquqi
witowwt
zxcjnc
oeit
potieq
iojge
nvoq
piqper

ans=8881647922834867244791415981705771412427494861672253136057167374729235842468240763290

1 1 1
a
a

ans=0

5 10 3
abcde
abc
bc
c
ans=1048576

Edited by author 05.04.2012 20:51

This is invalid test, but anyway

10
00:00:00
00:00:00
00:00:00
00:00:00
00:00:00
00:00:00
00:00:00
00:00:00
00:00:00
00:00:00

Answer:
00:13:58
00:13:58
00:13:58
00:13:58
00:13:58
00:27:56
00:27:56
00:27:56
00:27:56
00:27:56

#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