Could you help me?! :/

what can be in test 2/////

Have somebody any things about test N 26? Thank you!

Edited by author 18.02.2014 14:04

First of all, yes, the two players can be in the same cell, even at the beginning of the game. That only means that the first player that arrived there will take its value.
Test:
5
1 2 3 4 5
1 1
Answer:
15 0

Test:
5
1 2 3 4 5
5 5
Answer:
15 0

Why I have Runtime error? I use C#. Code is here.
using System;
using System.Globalization;

class EntryPoint
{
    static void Main()
    {
        CultureInfo ci = new CultureInfo("en-US");
        double p = Convert.ToDouble(Console.ReadLine(),ci);
        double q = Convert.ToDouble(Console.ReadLine(),ci);
        int x = 1;
        while (true)
        {
            if ((int)(q * x / 100) - (int)(p * x / 100) > 0)
                break;
            x++;
        }
        Console.WriteLine(x);
    }
}

try this test

input:
4 3 12
1 2 5
1 3 4
3 4 4

output:
YES

#include<iostream>
#include<math.h>
#include<algorithm>
void quickSort(double x[],double y[],int left, int right) {
      int i = left, j = right;
      double tmp;
      double pivot = x[(left + right) / 2];
      while (i <= j) {
            while (x[i] < pivot)
                  i++;
            while (x[j] > pivot)
                  j--;
            if (i <= j) {
                  tmp = x[i]; x[i] = x[j]; x[j] = tmp;
                  tmp = y[i]; y[i] = y[j]; y[j] = tmp;
                  i++;
                  j--;
            }
      };
      if (left < j)    quickSort(x,y,left, j);
      if (i < right)   quickSort(x,y,i, right);
}
double dis(double x1,double y1,double x2,double y2)
{
  return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
using namespace std;
int main()
{
   double a[305][305],x[305],y[305],max,s1;
   int i,j,n,k;
   cin>>n;
   for(i=1;i<=n;i++)cin>>x[i]>>y[i];
   quickSort(x,y,1,n);
   int left=1,right=1;
   for(i=2;i<=n;i++)
   {
      if(x[i]==x[i-1])right++;
      else
      {
          sort(y+left,y+right+1);
          right++;
          left=right;
      }
   }
   sort(y+left,y+right+1);
   for(i=1;i<n;i++)
   for(j=i+1;j<=n;j++)
   a[i][j]=dis(x[i],y[i],x[j],y[j]);
   for(i=1;i<n-1;i++)
   for(j=i+1;j<n;j++)
   {
     max=a[i][j];
     bool f=0;
     for(k=j+1;k<=n;k++)
     if ((x[k]-x[i])*(y[i]-y[j])==(y[k]-y[i])*(x[i]-x[j]))
     {
       if (max<a[i][k]) max=a[i][k];else a[i][k]=0;
       if (max<a[j][k]) max=a[j][k];else a[j][k]=0;
     }
     if (max>a[i][j]) a[i][j]=0;
   }
   s1=0;
   for(i=1;i<n;i++)
   for(j=i+1;j<=n;j++)s1+=a[i][j];
   s1=s1+0.4999999999;
   cout<<(long long)s1;
   system("pause");
   return 0;
}

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;

public class Solving {

    public StreamTokenizer in  = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    public PrintWriter out  = new PrintWriter(new OutputStreamWriter(System.out));

    public static void main(String[] args) throws IOException {
        new Solving().run();
    }

    public int nextInt() throws IOException {
        in.nextToken();
        return (int) in.nval;
    }

    public void run() throws IOException {
        int num = this.nextInt();
        int[] arr = new int[num];
        int a = 0;
        int b = 0;
        for (int iter1 = 0; iter1 < num; iter1++) {
            arr[iter1] = this.nextInt() + this.nextInt();
        }
        StringBuilder str = new StringBuilder();
        for (int iter2 = (arr.length - 1); iter2 > 0; iter2--) {
            a = arr[iter2]/10 + arr[iter2 - 1]%10;
            str.insert(0, (a%10 + b));
            b = a/10;
        }
        if (b != 0) {
            str.insert(0, b);
        }
        str.append(arr[arr.length - 1]%10);
        this.out(str.toString());
    }

    public void out(String str) {
        out.println(str);
        out.flush();
    }
}

I don't know how make this faster. Help me please.

Edited by author 16.02.2014 03:25

1) множитель не может быть отрицательным
2) умножать надо не по количеству восклицательных знаков а пока i*k < n
3) если ( n - i*k ) равно n % k или ( n - i*k ) равно k значит мы посчитали факториал ( то есть надо больше ничего с числом не делать )
4) n и k от 1 до 20
5) составители задачи тупые мудаки

This is the fragment of my program:
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
    int p;
    scanf("%d", &p);
    p--;
    if (p >= n)
        throw;

If I remove throw, I'll have WA 3.

So children can say absolutely incorrect information?

Here:

7S 9S KS 7C 9C QC 7D 8D 9D 7H
QS 10S AC KC 8C KD 10D AH 9H 8H
AS JS 8S JC 10C AD QD JD JH 10H
3
;(

7S 10S 7C 8C 9C JC 7D 9D 7H 8H
QS JS 8S KC QC 8C AD 8D QH 10H
AS KS 9S 10C QD JD 10D KH JH 9H
3
;(

7S 10S 7C 8C 9C JC 7D 9D 7H 8H
QS JS 8S QC 8C AD 8D QH 10H KH
AS KS 9S KC 10C QD JD 10D JH 9H
3
;)

7S 7C 7D 10D JD QD KD 7H 9H JH
10S 9S KC 9C 8C AD 9D 8D AH 8H
AS KS QS 8S AC JC 10C KH QH 10H
3
;(

7S 7C 7D 10D JD QD KD 7H 9H JH
10S 9S KC 9C 8C 9D 8D AH 8H KH
AS KS QS 8S AC JC 10C AD QH 10H
3
;(

7S 7C 7D 10D JD QD KD 7H 9H JH
8S 9S 10C 9C 8C AD 9D 8D 8H 10H
AS KS QS 10S AC KC JC AH KH QH
3
;)

7S 7C 7D 9D JD KD 7H 10H JH QH
AS AC KC QC JC 9C AD QD 10D KH
KS 10S 9S 8S 10C 8C 8D AH 9H 8H
2
;(

9S 10S JS 9C 10C JC 9D 10D JD JH
8S 7S 8C 7C 8D 7D 7H 8H 9H 10H
AS KS QS AC KC QC AD KD QD QH
2
;)

AC 7D KD QH 8H 8C 8S 9C QC 10C
KH 7H JH 7C 9H JS QD 8D 7S JD
10H KC AS AH QS JC 10D KS AD 10S
1
;)

7C AH QC KC KD 9C 9H 8C 10C JH
QS JC 9D 10H KH JD JS AC 8S AS
10D 7H KS 10S AD 7D 8H QH 8D QD
1
;(

8C 10S 9D 7S KD QC 9C KS JC 8S
8H JD 7D QH 10H 10D 8D 10C AH JS
KC QD 9S JH QS AS KH 9H 7H AC
3
;(

Edited by author 14.02.2014 01:31

Edited by author 14.02.2014 01:31

I have been working on it for a week....

Ребята будьте внимательный когда читаете условия. Я потратил пол часа пока понял что нужно составить не менее 6 а не не менее n слов...

Hello gentlemen!

Who can explain why I get WA8? Are there any test data available to anyone?

wrong answer, but I do't know why! help, please

#include<iostream>
using std::cout;
using std::cin;
using std:: endl;
int main(){
    int res = 0;
    int n1, n2, n3;
    cin >> n1;
    unsigned *arr1 = new unsigned[n1];


    unsigned temp[100];
    for(int i = 0; i < n1; ++i){
        cin >> arr1[i];
    }
    cin >> n2;
    unsigned *arr2 = new unsigned[n2];
    for(int i = 0; i < n2; ++i){
        cin >> arr2[i];
    }
    cin >> n3;
    unsigned *arr3 = new unsigned[n3];
    for(int i = 0; i < n3; ++i){
        cin >> arr3[i];
    }
    int sum = 0;
    for(int i = 0; i < n1; i++){
        for(int j = 0; j < n2; ++j){
            if(arr2[j] == arr1[i])
            temp[i]= arr2[j];
            sum++;
        }
    }
    int sum1 = 0;
    for(int i = 0; i < sum; i++){
        for(int j = 0; j < n3; ++j){
            if(arr3[j] == temp[i])
            sum1++;
        }
    }
cout << sum1 << endl;
    delete[]arr1;
    delete[]arr2;
    delete[]arr3;
system("PAUSE");
return 0;
}

I make matrices from the sequence of Catalan numbers 1,2,5,14,42,132,429,1430,4862,...
Such matrices satisfy all criterion:
1)All numbers are positive integers
2) Determinants of all matrices are equal to 1
But I get WA2

/*
 * File:   main.cpp
 * Author: Maxim Cherchuk <maxim.cherchuk@gmail.com>
 *
 * Created on 12 Февраль 2014 г., 23:36
 */


#include <iostream>
#include <algorithm>

using namespace std;

/*
 *
 */

const int MAXN = 50000, MAX = 10000;
int a[MAXN], b[MAXN];
int n, k, m;

int main(int argc, char** argv) {
    ios_base::sync_with_stdio(0);
    cin >> n;
    for(int i = 0; i < n; ++i) {
        cin >> a[i];
    }
    cin >> k;
    for(int i = 0; i < k; ++i) {
        cin >> b[i];
    }
    sort(a, a+n);
    sort(b, b+k);
    for(int i = 0; i < n; ++i) {
        const int cur = a[i];
        int l = 0;
        int r = k;
        while(r-l > 1) { // binary search
            m = (l+r)>>1;
            if(cur+b[m] > MAX)
                r = m;
            else
                l = m;
        }
        if(cur + b[l] == MAX) {
            cout << "YES";
            return 0;
        }
    }
    cout << "NO";
}