Hey guys any hints here? All examples from forum works on my program but this one
exceeds. I guess it's because more than 200 characters are written, but I cannot find any test case that will require less written characters but the output will have more length

I use a sement tree,  gave tle in test 17 and when I changed the scanf by the following "std :: ios :: sync_with_stdio (false)" I have acepted with time of 0.358 s

Since there can be more than one acceptable solution, I think, it could give a more unambiguous result.

who could give some testdata to me ?


3  3
9
10
11

result?

Edited by author 05.11.2008 04:11

using System;

class Program
{
    static void Main()
    {
        int t = int.Parse(Console.ReadLine());
        string[] s = Console.ReadLine().Split(' ');
        Array.Sort(s);
        int r = 0;

        for (int i = 0; i < (t / 2)+1; i++)
        {
            r += ((int.Parse(s[i]) / 2) +1);
        }
        Console.WriteLine(r);
    }
}

Edited by author 16.06.2012 19:10

Edited by author 16.06.2012 19:11

Hi, folks!

Can anyone give me the source code in C# of this task (mine is already AC) when the timing is less than 0.1 sec?
Or can anyone give me the key notes how it can be decreased in almost ten times?

Thanks in advance!

#include <iostream>
using namespace std;

int main() {
    int cand, ch;
    cin >> cand >> ch;
    int* Ch = new int[ch];
    double* Cand = new double[cand];

    for (int i = 0; i < cand; i++) {
        Cand[i] = 0;
    }

    for (int i = 0; i < ch; i++)
        cin >> Ch[i];

    for (int i = 0; i < ch; i++) {
            Cand[Ch[i]-1] += 1;
    }

    for (int i = 0; i < cand; i++) {
        printf("%.2f%%\n", (Cand[i]/6.0)*100.0);
    }
}
+++++++++++
Why this isn't work?

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

int main() {
    int i,j,N,p;
    std::cin>>N;

    float cd[N][2],R,dist = 0.0;
    std::cin>>R;

    for(p = 0;p < N;p++)
        std::cin>>cd[p][0]>>cd[p][1];

    i = 0;
    j = i + 1;

        dist = (2 * std::acos(-1.0) * R);

    while(i < N)
    {
        dist += std::sqrt( ( (cd[j][0] - cd[i][0]) * (cd[j][0] - cd[i][0]) ) +
( (cd[j][1] - cd[i][1]) * (cd[j][1] - cd[i][1]) ) );
        i++;
        j = (i + 1) % N;
    }

    std::cout <<std::setprecision(2) << std::fixed <<dist;
    return 0;
}

What answer should be for given test?

1 1 1 1
0

Can anyone explain the sample:

3
0 0 0
1 1 0
1 1 0



1
1 3


I dont understand it. The resulting matrix is:
0 1 1
0 1 1
0 0 0

Thanks!

Same solution:

Visual C++ 2013:
Accepted    0.686    8 120 КБ

G++ 4.9 C++11:
Time limit exceeded    21    1.513    3 924 КБ

oh my code..

first  use dynamic programming to find the minimum value of distance from 0 and visit 1~m once then to 0, we can get a chain 0--i1-i2--...-im--0

then we consider to cut this chain, use dynameic programming dp[i] to record distance.we cut first i node of the chain..
dp[m] is the result..

I'll try this idea...

rt

can someone explain test number 2?

please post it (read chunk) here or give some piece of code, many sanx!

#include <stdio.h>
#include <math.h>

int main( int argc, char ** argv )
{
    unsigned long number[256*1024];
    int counter = 0;
    while( scanf("%lu",&number[counter]) != EOF )
    {
        counter++;
    }

    for( counter--; counter >= 0; counter-- )
        printf("%.4f\n",sqrt(number[counter]));
    return 0;
}

i just found no of storngly connected components and got ac in 0.562 sec after eliminating stack overflow with pragma.

how to make it so fast as 0.031 sec and so little memory?

Try this test:

Input:
5 1
100 0
100 0
100 0
100 0
100 0
2 50

Output:
250

Hello! I have strange WA 22 and really don't understand why. Can anyone help to find the bug in my code. (corrected)

Edited by author 04.12.2016 20:25

import java.util.*;
import java.io.*;

public class MEGA
{
    public static void main(String arg[])
    {
        int k,n,i,sum=0;
        Scanner scan=new Scanner(System.in);
        k=scan.nextInt();
        n=scan.nextInt();
        int[] a=new int[n];
        for(i=0;i<n;)
        {
            if(scan.hasNextInt())
            {
                a[i]=scan.nextInt();
                i++;
            }
        }
        for(i=0;i<n;i++)
        {
            sum+=a[i]-k;
        }
        if(sum<0)
        {
            System.out.println("1");
        }
        else
        {
            System.out.println(""+sum);
        }
    }
}

This very simple program is turning out to be a headache. What is Test #3?