using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Text;


namespace Пример
{

    class Programm
   {
         static void Main(string[] args)
        {
           int num= int.Parse( Console.ReadLine());

            int sum = 0;
            if (num<10000&&num>-10000)
            {


                if (num > 0)
                {
                    for (int i = 1; i <= num; i++)
                    {
                        sum += i;
                    }
                }
                else if (num < 0)
                {
                    for (int i = num; i <= 1; i++)
                    {
                        sum += i;
                    }
                }
            }

            Console.WriteLine(sum);
        }


    }

}

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Text;


namespace Пример
{

    class Programm
   {
         static void Main(string[] args)
        {
           int num= Convert.ToInt32( Console.ReadLine());

            int sum = 0;
            if (num > 0)
            {
                for (int i = 1; i <= num; i++)
                {
                    sum += i;
                }
            }
            else if (num < 0)
            {
                for (int i =num; i <=1; i++)
                {
                    sum += i;
                }
            }

            Console.WriteLine(sum);
        }


    }

}

There is an undirected graph with n vertices and m edges. There have been three types of requests: to add an edge (X_i, y_i), remove the edge (X_i, y_i), and check whether or not vertices x_i and y_i connected.

I read  this problem(disjoint set union division) in this site http://e-maxx.ru/algo/sqrt_decomposition (this is in russian but i use google translate)
The approach is also given, they use disjoint sets with sqrt decomposition. I didn't understand this . plz help.

My algo is right. Now i have AC. My problem is input string. I rewrite my program on Pascal (i use ANSISTRING) and got AC.

Edited by author 21.08.2016 18:31

помогите, что в нём?!?!


спасибо, за помощь )

#include <iostream>
#include <math.h>
#include <algorithm>
#include <vector>

using namespace std;

int main()
{
    vector<long long int>v;

    long long int b;

    while (cin >> b) v.push_back(b);

    reverse(v.begin(),v.end());

    for (long long int k: v) cout << fixed << sqrt(k) << "\n";

    return 0;
}

The problem's author did not write that, after the last move you need to print out a position, but it follows from the third example and tests
Please add this condition

My code is ok all tests before #8, but I've WA on #8... Anything that I don't see?

Thanks

No subject

Edited by author 13.12.2017 23:34

This problem involves number theory???

#include <iostream>

using namespace std;

int main()
{
    int k;
    cin >> k;

    int n;
    cin >> n;

    int change;
    int result = 0;

    for (int i = 0; i < n; i++)
    {
        cin >> change;
        result += change;
    }

    if (result - k*n >= 0) cout << result - k*n;
    else cout << 0;
    return 0;
}

Насколько я понял, есть орграф без циклов, при этом если из вершины v1 можно попасть в v2, из v2 в v3, то и из v1 можно попасть в v3 (дуга моделирует победу). Команда v1 лучше v2, если нет u, отличной от v1 и v2/ из u можно попасть в v1 и v2.

Код с WA2:
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>

const int nmax = 1005;

bool a[nmax][nmax];
int d[nmax], p[nmax];  //d[v]=0,если нет входящих дуг в v, p[v]-вершина u/ v достижима из u и d[v]=0.
bool used[nmax];
char ch;

int n, q, v, u;

void DFS(int v, int par);

void main()
{
    scanf("%d\n", &n);
    for (int i = 1; i <= n; i++)
    {
        d[i] = 0;
        used[i] = false;
    }

    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            scanf("%c", &ch);
            if (ch == '0') a[i][j] = 0;
            else a[i][j] = 1;
            d[j]+=a[i][j];
            if (j == n) scanf("\n");
        }
    }

    for (int i = 1; i <= n; i++)
    {
        if (d[i] == 0)
        {
            p[i] = i;
            DFS(i, i);
        }
    }

    scanf("%d\n", &q);
    for (int i = 1; i <= q; i++)
    {
        scanf("%d %d\n", &v, &u);
        if ((d[v] > 0) && (d[u] > 0) && (p[v] == p[u])) printf("No\n");
        else printf("YES\n");
    }
}

void DFS(int v, int par)
{
    used[v] = true;
    for (int i = 1; i <= n; i++)
    {
        if ((!used[i]) && (a[v][i] == 1))
        {
            p[i] = par;
            DFS(i, par);
        }
    }
}

Ошибка была в том, что из несколько вершин, не имеющих входные дуги, можно попасть в некоторую вершину.

Edited by author 03.01.2017 22:31

31
1 2 2 3 3 3 3 3 3 4 4

28
1 2 2 2 3 3 3 3 3 3 3

26
1 1 1 1 1 1 2 3 4 5 6

Edited by author 16.08.2016 04:04

Very easy:)
just print numbers 1,2,...,n
then n+1,2*n+1,...,m*n+1

input:
2 3
Output:
1 2
3 5 7

Why this result failed 1 test?

If anybody knows O(n) solution, please, send it to sk1@hotbox.ru
PS: My solution is O(n^2)

I don't understand my mistake. Please, give me some tests to solve this problem.

That's my q.

Should I use some test of simplicity (from Cormen or somewhere else) or is there any better way for such numbers?