#include <bits/stdc++.h>

using namespace std;
struct duy{
    long long x,y;
};
duy a[100005];
bool cmp1(duy a, duy b){
    return a.x>b.x;
}
bool cmp2(duy a, duy b){
    return a.y>b.y;
}
long long n,dem=0,t=-99,s[100005][2],duoi[100005],res=0;
double q;
int main()
{
    cin>>n;
    for(long long i=1; i<=n; i++) cin>>a[i].x>>a[i].y;
    sort(a+1,a+n+1,cmp1);
    for(long long i=1; i<=n; i++){
        if(a[i].x==t) s[dem][2]++;
        else{
            dem++;
            s[dem][1]=a[i].x;
            s[dem][2]=1;
            t=a[i].x;
        }
    }
    duoi[dem]=0;
    for(long long i=dem-1; i>=1; i--) duoi[i]=duoi[i+1]+s[i+1][2];
    for(long long i=1; i<=dem-1; i++) res=res+s[i][2]*duoi[i]*(s[i][1]-s[i+1][1]);
    sort(a+1,a+n+1,cmp2);
    dem=0;
    for(long long i=1; i<=n; i++){
        if(a[i].y==t) s[dem][2]++;
        else{
            dem++;
            s[dem][1]=a[i].y;
            s[dem][2]=1;
            t=a[i].y;
        }
    }
    duoi[dem]=0;
    for(long long i=dem-1; i>=1; i--) duoi[i]=duoi[i+1]+s[i+1][2];
    for(long long i=1; i<=dem-1; i++) res=res+s[i][2]*duoi[i]*(s[i][1]-s[i+1][1]);
    q=((double)((double)2*(double)res))/((double)((double)n*(double)(n-1)));
    cout<<roundf(q*1)/1;
}

var b:array[1..4] of int64; i:integer;
begin
 for i:=1 to 4 do
 read(b[i]);
 for i:=4 downto 1 do
 writeln(sqrt(b[i]):0:4);
end.

I got TLE, but when i used Visual C++ 2010 I hot AC. It worked for me, I hope it will work also for you

WBBB
BBBW
WBWB
BBBB
101
010
101

2



BBBB
BBWW
BBBB
WWBW
010
111
010

3

I use priority queue,the memory that I used is 236 KB,but I still WA at test 5,but I don't know why...

i use seive but it still verdict TLE
my code https://ideone.com/MV23r8

I am using bfs to solve the problem but still getting TLE. Can anyone suugest the cause?

My graph consists of vertexes, which are points, that allow diagonal crossing. Edge (a, b) exists if and only if (a) strictly south-west from (b). Since graph is built, simple dfs easily helps solve the problem ( length of the longest path in the graph - it's diagonal part of result path ).
To avoid TLE#10 in this approach, memorization should be used.

I know, it's not the best approach, but since constraints are relatively low, it does work.
My AC - 0.015    164 КБ.
I think, Longest increasing subsequence - is better in general case.

I got AC. My time is 0.001s (This problem #1222) - Objected-oriented progarmming, call methods, cycles in cycles...

But my time of Problem #1000 (A+B) (int + int) - 0.015s

Why?

:)))

Доброго всем времени суток.
Прошу помочь разобраться, в чем может состоять проблема моего кода. В NetBeans все компилируется без проблем, здесь же выдается ошибка RuntimeError. Я здесь впервые, поэтому прошу не судить (З.Ы. Руководство по написанию java скриптов прочел и до сих пор не нашел ошибки). Заранее всем спасибо.

Hi all!
I'm asking you for help, cause my script cannot be run by server java compiler. In NetBeans it works well, but here that's a RuntimeError.
Thank you in anvance.

import java.util.Scanner;
import java.io.PrintWriter;

public class Cryptography {

    final String alphabetStr = "abcdefghijklmnopqrstuvwxyz";

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

    void run() {
        Scanner in = new Scanner(System.in, "ISO-8859-1");
    PrintWriter out = new PrintWriter(System.out);

    String word = in.next();

        out.println( descript(word) );
    out.flush();
    }

    String descript(String word) {
        int index = alphabetStr.indexOf( word.charAt( word.length()-1 ) );

        if ( word.length()==1 ) return ("" + alphabetStr.charAt(index-5));

        index -= alphabetStr.indexOf( word.charAt( word.length()-2 ) );
        if ( index<0 ) index += 26;
        return descript(word.substring(0, word.length()-1)) + alphabetStr.charAt(index);
    }

}

Edited by author 17.02.2017 16:13

Edited by author 17.02.2017 16:15

This is my program:

#include <stdio.h>
int len[16];
int ldist[16];
int rdist[16];
int depart;
int gor,gol,iv;
int N;
void prelr()
{
    int i;
    ldist[0]=0;
    for (i=1;i<N;i++)
        ldist[i]=ldist[i-1]+len[i-1];
    for (i=N-2;i>=0;i--)
        rdist[i]=rdist[i+1]+len[i];
}
void read()
{
    scanf("%d",&N);
    int i;
    for (i=0;i<N-1;i++)
        scanf("%d",&len[i]);
    scanf("%d",&depart);
    depart--;
    scanf("%d%d%d",&iv,&gor,&gol);
}
int best;
bool first;
int deptime;
void nexttimepoint(int&time,int&posi,int&dir)
{
    //Test nearest shuttle
    int tmp;
    if (dir==1)
    {
        tmp=((time-ldist[posi]-gor)%iv+iv)%iv;
        if (tmp!=0) tmp=iv-tmp;
        time+=tmp;
    }
    else
    {
        tmp=((time-rdist[posi]-gol)%iv+iv)%iv;
        if (tmp!=0) tmp=iv-tmp;
        time+=tmp;
    }
    //Record if it's first
    if (first)
    {
        first=false;
        deptime=time;
    }

    //Make path
    if (dir==1)
    {
        time+=len[posi];
        posi++;
    }
    else
    {
        time+=len[posi-1];
        posi--;
    }

    //Test if reverse needed
    if (posi==0)
        dir=1;
    if (posi==N-1)
        dir=-1;

}
void testdepart(int direct)
{
    int i,time,p,delayi,dir;
    int last;
    for (i=1<<(N-2);i-->0;)
    {
        delayi=i;
        last=(1<<N)-1;
        if (direct==1)
            time=ldist[depart];
        else
            time=rdist[depart];
        p=i;
        dir=direct;
        p=depart;
        first=true;
        nexttimepoint(time,p,dir);
        while (p!=depart||dir!=direct)
        {
            if (p==0||p==N-1)
                time++;
            else if (last&(1<<p)&&p!=depart)
            {
                if (delayi&(1<<(p-1)))
                    delayi^=1<<(p-1);
                else
                {
                    last^=1<<p;
                    time++;
                }
            }
            nexttimepoint(time,p,dir);
        }
        if (best>time-deptime)
            best=time-deptime;
    }
}
void process()
{
    prelr();
    if (depart!=N-1)
        testdepart(1);
    if (depart!=0)
        testdepart(-1);
}
int main()
{
    read();
    best=2147483647;
    if (N==1)
        printf("0\n");
    else
        process();
    printf("%d\n",best);
    return 0;
}

Thanks.

using System;

namespace ConsoleApplication
{
    internal class Program
    {
        public static void Main(string[] args)
        {
            var i = Convert.ToInt32(Console.ReadLine());
            var v = i;
            var b = v;
            var z = Math.Abs(i);
            var x = 0;
            if (b<0)
            {
                x = -2;
            }
            else
            {
                x = 0;
            }
            var sum = 0;
            while (x!=z)
            {
                sum = sum + v;
                if (b<0)
                {
                    v++;
                }
                else
                {
                    v--;
                }
                x++;
            }
            Console.WriteLine(sum);
        }
    }
}

The primality of the table for n=5 is equal to 3 ?

[code deleted]

Edited by moderator 19.11.2019 23:54

#include<iostream>
#include<math.h>
using namespace std;

int main()
{
    int n;
    int p;
    double a;
    int i;
    cin>>n;
    int flag;
     for(p=sqrt(n*2);p>0;p--)
     {
         a=(double)n/p-(double)(p-1)/2;
        if(a==(int)a && a>0)
        {
            cout<<a<<" "<<p;
            return 0;
        }

     }
    cout<<n<<" "<<1;
    return 0;
}

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

#define SIZE 1024

int main(void)
{
    unsigned long long int *buffer = \
    (unsigned long long int *) malloc(sizeof(unsigned long long int) * SIZE);
    int i = 0;

    while (scanf("%llu", &buffer[i]) > 0)
        i++;

    for(i -= 1; i >= 0; i--)
        printf("%.4f\n", sqrt(buffer[i]));

    return 0;
}

У меня всё работает, на тестовых данных выдаёт правильный результат. В чём косяк?

Edited by author 14.02.2017 15:27

thanks

#include <bits/stdc++.h>

using namespace std;
float p,q,t1,t2,t3,l1,l2;
int main()
{
    cin>>p>>q;
    for(int i=1; i<=1000000; i++){
        t1=(p*(float)i)/(float)100;
        t2=(q*(float)i)/(float)100;
        l1=(int)t1;
        l2=(int)t2;
        if(l1==t1||l2==t2||l1!=l2){
            if(l1!=t1) l1=l1+1;
            for(int j=l1; j<=l2; j++){
                t3=((float)j/(float)i)*(float)100;
                if(p-0.000001<t3&&t3<q+0.000001){
                    cout<<i;
                    return 0;
                }
            }
        }
    }
}

I can only get

2 3 1
3 4 3
2 3 2
according to the sample output...

has anyone done it in O(n)?