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

Let's sequence S(n) = {s[1], s[2], ..., s[n]} is optimal sequence for the first n steps. L[i] is max number when we can get one space using S(n) for any number in range [1..L[i]]: firstly, we apply s[n], then s[n-1] and etc.
Now let's get next s[n+1]. Suppose we found s[n+1] and want get L[n+1]. Always L[n+1] + 1 = q*s[n+1] + r, 0<=r<s[n+1]. L[n+1] is max number => for L[n+1] + 1 we need at least n+2 steps => q + r >= L[n] + 1. But L[n+1] + 1 is min number when using S(n+1) we can't get one space => r->max, q->min =>
r = s[n+1] - 1, q = L[n] + 1 - r =>
L[n+1] + 1 = q * s[n+1] + r =>
L[n+1] = (L[n] - s[n+1] + 2) * s[n+1] + s[n+1] - 2 -> max (where arg is s[n+1])
=> s[n+1] = (L[n] + 1) / 2
But using symmetry we can give s[n+1] = floor(L[n] / 2) + 1 (we choice bigger number because s[2] > 1).
=> L[n+1] = q * s[n+1] + r - 1, q = L[n] + 1 - r, r = s[n+1] - 1.



Edited by author 26.02.2022 01:27

Edited by author 26.02.2022 01:27

My program runs any tests from other topics perfectly, but something is wrong with the task 1. Sounds like madness.

Edited by author 24.02.2022 23:04

Edited by author 24.02.2022 23:05

Pay attention, that points given in order A, B, C. That means that angle of your triangle must equal to angle BAC

Edited by author 21.02.2022 17:07

Set types as long long for variables used as sum of cables lengths

You should write "fixed<<setprecision(15)", because you get uncorrected format of numbers with dot

There's negative numbers, that's mean if you're solving problem with C++, remainder will be calculating wrong

Определим переменную difference
Затем для каждого элемента истории прибавляем к difference разницу между
числом с компьютера - 2 и введённым числом

В конце проверяем k - 2 + difference, если оно меньше нуля, то мы в пролете, иначе вывести число

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main()
{
    double val = 0;
    vector<double> v;
    char ch = 0;

    while (cin >> val)
    {
        v.push_back(val);
    }
    for (int i = 0; i < v.size(); i++)
    {
        cout << sqrt(v[i]) << endl;
    }
}

I solved this problem with algo of Dijkstra.
Edge : (i,j)->(i+1,j), (i-1, j), (i, j+1) with weight of a[i]j];

First DP formula like 1 + min max (a[i - 1, k-1], a[n - i,k ]) is NOT GOOD! In pascal it gets TLE!
Take more clever formula!!!

I hope that all tests take into account the multiplicative function condition: F(1) == G(1) == 1

what are mistakes in this code?
#include <iostream>
using namespace std;
long long int a[15000];
long long int b[1000000];
int main()
{
    int c, f, d, e, g,l;
    l = 0;
    cin >> c;
    for (f = 0; f != c; f++)
    {
        cin >> a[f];
    }
    cin >> d;
    for (e = 0; e != d; e++)
    {
        cin >> b[e];
    }
    for (e = 0; e != d; e++)
    {
        g = b[e];
        for (f = 0; f != c; f++)
        {
            if (g == a[f])
            {
                l = l++;
                break;
            }

        }
    }
    cout << l << endl;
    system("pause");
    return 0;
}

Edited by author 18.02.2022 18:53

После того как сайт проверил мою задачу, я решил проверить несколько частных случаев
и в одном у меня программа ломалась, но это не помешало ей пройти проверку.
Входные данные:
5 4
1 2
2 3
3 4
4 5
Ответ 2 и 23 и 45, но должно быть 2 и 12 и 45

#include <iostream>
using namespace std;

int main() {
unsigned long long n,a,b,c,s;
int i;
unsigned long long  x[100001];
memset(x,0,sizeof(x));
cin>>n;
for (i=1;i<=n+1;i++)
{
cin>>a>>b>>c;
x[a]=x[a]+c;
x[b+1]=x[b+1]-c;
};s=0;
for (i=1;i<=n;i++)
{
s=s+x[i];
if (i==n) cout<<s<<endl; else
cout<<s<<" ";
}
};

Try std::unordered_map instead of std::map. This one helped me

Where is my mistake?
I have WA on test 2!!!

#include <iostream>
using namespace std;
int main ()
{
    char c;
    int i,n,br=0,b=0,l=1;
    while (cin>>c)
    {
        if (c=='!'||c=='?'||c=='.') l=1;
        else
        {
            if (c>='a'&&c<='z')
            if (l==1) br++;
            if (c>='A'&&c<='Z')
            if (l==0) br++;
            l=0;
        }
        if (c=='!'||c=='?'||c=='.')
        {
            if (br>0) b+=1;
            br=0;
        }
    }
    cout<<b<<endl;
    return 0;
}

Spend 10 submissions to realise, that I don't return anything from function, that must return bool value

Suppose we want to calculate max({A * a[i] + B * a[i + 1], A * a[i + 1] + B * a[i]}) for i = 0..n-1 (in this problem A = 0, B = 1). Then answer(A, B, n) = max(answer(max(A, B), A + B, n/2), A * a[n - 1] + B * a[n], B * a[n - 1] + A * a[n]), so it can be solved recursively.