I'm native Russian language speaker, but I really cannot understand what is required in the problem:

"Поскольку Вадим не любит сладкое, в конце он выберет себе самый маленький по площади кусочек, однако он не хочет, чтобы кто-то это заметил. Для этого современному обер-форшнейдеру нужно подобрать такой способ элегантно разделить торт, чтобы часть, которая достанется ему, была наибольшего размера."

He doesn't like sweets, so he takes part of minimal size, but he wants to maximize size at the same time. WTF, where is the logic here, what does author want from us - to find maximin over all cuts, or something else?

My program worked correctly on N in 1 to 9, and incorrectly on numbers greater than 9 (error in string concatenation). I kept getting an error on test 2 until I fixed the error. Apparently, in test 2, the input is a number greater than 10.
Try this test:
in:
11
out:
((((((((((sin(1)+11)sin(1-sin(2))+10)sin(1-sin(2+sin(3)))+9)sin(1-sin(2+sin(3-sin(4))))+8)sin(1-sin(2+sin(3-sin(4+sin(5)))))+7)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6))))))+6)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7)))))))+5)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7-sin(8))))))))+4)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7-sin(8+sin(9)))))))))+3)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7-sin(8+sin(9-sin(10))))))))))+2)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7-sin(8+sin(9-sin(10+sin(11)))))))))))+1

I made all my coordinate variables double, so I had a bug when I printed it out :(
cout << (int)x << ' ' << int(y) << '\n'; -> fixed my bug

I have tested, and 500 kbytes input reads about 20-30 ms in acm.timus.ru system.
It equal to ~25 Mbytes/s speed read.

Seems there installed HDD disk.

There any plan to use fastest SSD disks in server side ?

Many problems solutions actually took many times for accessing input/output, not calculating actual algorithm.

6 11 1 1
1 2
1 3
3 1
3 2
2 6
2 4
3 5
5 5
5 4
4 6
6 4

Answer: 0

Just prepare the equation as described and solve for c=x,

display_sum + k - (entered_sum + x) == (n + 1) * 2;
display_sum + k - entered_sum - x = 2 * n + 2;
x = -2 * n - 2 + display_sum + k - entered_sum;

if x < 0 then Big Bang!

56 2
56 68

49 4
7 41 43 65

If you get WA5, check case, where policeman catch the thief until first tram arrive.

I tried many times and I got TLE7, on code with cin cout, and the visual studio compiler. BUT with this input only 5 million numbers. It should not give TLE (I used ios and tie), which is more strange in g++, ABSOLUTELY the same code gives AC, with a time of 0.187, the code is MORE THAN 5 TIMES FASTER! I understand that g++ and visual differ a lot in terms of compiler settings, but to have such a big difference, I'm just shocked. I am very interested to find out why this is happening. Usually visual is faster, it is very interesting to find out which settings and in which cases make the compiler faster.

you don't understand what your code does at all
try to think more

#include <bits/stdc++.h>

using namespace std;

#define go ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
typedef long long ll;

int main() {
    go
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    while(a < c){
        if(a+b > c) break;
        a += b;
        c -= d;
    }
    cout << max(a, c);

    return 0;
}

Let's assume delta is the difference between the nearest number in our set. Then we can approve delta will decrease by half (at least) in every 2 rounds. So the simulation will be executed at most 2log 10^18 time, it's about 120.

k = 1
k = 2^m for some m
k = n
k = n-1

cases are enough for solve problem :)

#include <iostream>
#include <vector>
using namespace std;
int main()
{
    int N = 45;
    int M;
    cin >> M;

    vector <long long> a(N+2);
    a[0] = 2, a[1] = 2;



    for (int i = 0; i < N; i++)
    {
        a[i+2] = (a[i+1] + a[i]);
    }



       //Fibonacci sequence

       // F0 = 2;
       // F1 = 2;
       // Fn = Fn-1 + Fn-2


    cout << a[M-1];
    return 0;
}

Problems 2158-2173 from Ural School Programming Contests 2021 and 2022 were added to the Problem set.

This is my tests:

<Test 1>
input:
10 5
8
1 0
2 0
3 1
4 1
5 0
6 0
15 1
16 1
output:
No reason
5 15
6 16
1 3
2 4

<Test 2>
input:
11 1
18
1 0
2 0
3 0
4 0
6 0
7 0
8 0
9 0
11 0
12 1
13 1
14 1
15 1
16 1
17 1
18 1
19 1
20 1
output:
Liar

<Test 3>
input:
4 2
8
7 0
8 0
9 0
19 1
20 0
21 1
22 1
23 1
output:
No reason
20 21
7 19
8 23
9 22

Edited by author 29.03.2023 21:01

try think like dp[length of string][sum of string] = count of numbers, answer is dp[9][s].

ds:=sqr(b)-a*c;

the checking must look like this
if ds+0.000000000001>=0 then
...

THIS TEST
6 5
1 3
6 3
11112
22221
11112
22212
21112
22222

ANS:7 1