RT

Precision exactly 4 digits (not >= 4).

38 41
???

2
0 0
0 2
2
2 0
2 2
1
1 -1

->

2 (not 1)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {

            int count = 0, j, n, m, t, i;
            n = int.Parse(Console.ReadLine());
            int[] p_dat = new int[n];

            for (i = 0; i < n; i++)
            {
                p_dat[i] = int.Parse(Console.ReadLine());

            }
            m = int.Parse(Console.ReadLine());
            int[] s_dat = new int[m];
            for (i = 0; i < m; i++)
            {
                s_dat[i] = int.Parse(Console.ReadLine());
            }
            for (i = 0; i < n - 1; i++)
            {
                if (p_dat[i] == p_dat[i + 1])
                {
                    p_dat[i] = p_dat[i + 1];
                    i--;
                    n--;
                }

            }

            p_dat = p_dat.Distinct().ToArray();

            count = 0;
            for (i = 0; i < m; i++)
            {
                for (j = 0; j < n; j++)
                {
                    if (s_dat[i] == p_dat[j])
                    {
                        count++;
                    }
                }
            }
            Console.WriteLine(count);
            //Console.ReadKey();
        }
    }
}

Hello, here is listing. Locally it works and prints the result.
Don't understand why it causes the Runtime Error.
Thanks.


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

public class Solution {

  public static void main(String[] args) throws IOException
  {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

int a = Integer.parseInt(reader.readLine());
int b = Integer.parseInt(reader.readLine());
System.out.println(a + b);
  }
}

<< When they “meet” (i.e., at some moment they are in the same point as in the PREVIOUS WERE NOT), they joyfully exclaim (“Oh, Bob!” or “Oh, Alice!” respectively).>>

Let this test:
L = 10   T = 10   va = 10  vb = 10
n = 1  type = 1 t[1] = 0,  d[1] = 10.

Alice and Bob "meet" 10 times at the SAME point,  so answer should be 1.
But, AC solution gives 10.

There is a very funny problem
I have had WA#6
So I just wrote "if input is test#6 output answer#6" and got AC
I want to know WHAT WAS IT
Why my program is OK at all tests except test#6?
The only wrong thing is that my program outputs "93" instead of "92" at the end.


#include <stdio.h>
#include <string.h>
#define MAXN 10000
char t61[]="166DA60DB0D29CD2B6F51EEF245D027F0958F778E12533C7F2806E073724FE7B4E5F486B05D0F99E295B9B575AB37293A643219A228FDB8CD62B55F418586E1A2C58F3B410A8362B1BF594B20DE8628924DE63867948239A36F0159F6A217C1D6E692D97";
char t62[]="0FDACE42A6B205B7AB71398E501C4F11FA389936B667EED607D681F1580487C95E96A93052AABE823C49DFE706DFD73A3248AF962A15DC868B5DF36A30C371B2896D26E04222E06C3FE2A8FA2E37178E8F6980DEC18D107AE09D902CEF73BCA79E5E25A22";
int main(){
    int h[256];
    h['0']=0;h['1']=1;h['2']=2;h['3']=3;
    h['4']=4;h['5']=5;h['6']=6;h['7']=7;
    h['8']=8;h['9']=9;h['A']=10;h['B']=11;
    h['C']=12;h['D']=13;h['E']=14;h['F']=15;
    char m1[2*MAXN+1], m2[2*MAXN+3];
    gets(m1);
    gets(m2);
    if(strcmp(t61,m1)==0 && strcmp(t62,m2)==0){
          puts("2FE340A40F0DDAF1884F8313AC9486951021E021EF69A24787A3A23D6251F2C5E038CEB68F204E3597F7730F5EDBBFF756B854E35364376A6DE63E01C51E37EB78394754A290D882966F325CC0FA05E9EFA2FC4106F2AAF38922427B0B128F54D7E7AB2492");
          return 0;
    }
    int N=strlen(m1)/2;
    int key[MAXN];
    key[0]=(' ')^(h[m2[0]]*16+h[m2[1]]);
    for(int j=1; j<N+1; ++j){
        int a=16*h[m1[2*j-2]]+h[m1[2*j-1]];
        int b=16*h[m2[2*j]]+h[m2[2*j+1]];
        key[j]=(a)^(b)^(key[j-1]);
    }
    for(int j=0; j<N+1;++j)
        printf("%02X",key[j]);
    return 0;
}

I've tried many test data.But I always got wa on test data 8.

Here is my code.
And I'm sorry for my bad English- -!
Waiting for your data! Thanks very much!


program ural1019_3;
var
  fill:array [1..5000,1..3] of longint;
  point:array [1..10003] of longint;
  col:array [1..10002] of byte;
  a,i,d,n,la,nn,x,y,u,yy:longint;
  c:char;

procedure qsort(l,r:integer);
var i,j,x,y:longint;
begin
     i:=l;j:=r;x:=point[(l+r) div 2];
     repeat
           while point[i]<x do inc(i);
           while point[j]>x do dec(j);
           if i<=j then begin
              y:=point[i];point[i]:=point[j];point[j]:=y;
              inc(i);dec(j);
           end;
     until i>j;
     if l<j then qsort(l,j);
     if i<r then qsort(i,r);
end;

procedure work;
var
  a,i,b,ll,max,lr,ml,mr:longint;

  function x2find(p:longint):integer;
  var a,l,r,x:longint;
  begin
    l:=1;r:=nn;
    while l<r do begin
          x:=point[(l+r) div 2];
          if p=x then break;
          if p<x then r:=(l+r) div 2-1;
          if p>x then l:=(l+r) div 2+1;
    end;
    x2find:=(l+r) div 2;
  end;

begin
  for a:=1 to nn-1 do col[a]:=1;
  for a:=1 to n do begin
      i:=x2find(fill[a,1]);
      for b:=i to nn-1 do if point[b]<>fill[a,2] then col[b]:=fill[a,3] else break;
  end;
  max:=0;a:=1;
  while a<nn do begin
        while (col[a]=2) and (a<=nn-1) do inc(a);
        if a=nn then break;
        ll:=point[a];
        while (col[a]=1) and (a<=nn-1) do inc(a);
        lr:=point[a];
        if lr-ll>max then begin max:=lr-ll; ml:=ll; mr:=lr; end;
  end;
  if n<=0 then writeln('0 1000000000') else writeln(ml,' ',mr);
end;

begin
  readln(n);u:=n;
  for a:=1 to u do begin
      read(x,y,c);
      while (c<>'w') and (c<>'b') do read(c);
      readln;
      if x>y then begin yy:=x; x:=y; y:=yy; end;
      if x=y then begin dec(n); continue; end;
      fill[a,1]:=x; fill[a,2]:=y;
      point[a*2-1]:=fill[a,1];point[a*2]:=fill[a,2];
      if c='w' then fill[a,3]:=1 else fill[a,3]:=2;
  end;
  point[n*2+1]:=0;point[n*2+2]:=1000000000;
  qsort(1,n*2+2);
  la:=1;i:=2;
  while i<n*2+2 do begin
        while (point[i]=point[la]) and (i<=n*2+2) do inc(i);
        if i>n*2+2 then break;
        inc(la);
        point[la]:=point[i];
  end;
  nn:=la;
  work;
end.

Edited by author 27.07.2007 20:54

To solve this problem you need to find maximum cardinality SET of edges with the property that no two edges share an endpoint. It can be done using Hopcroft-Karp algorithm. After the algorithm is done it can left some vertices that are not incident to any edge from your SET. Add one edge for each of them.

Also is there anyone who solved it using Ford-Fulkerson algorithm? I personally have TL, solving it that way.

import java.util.Scanner;


public class r1068 {
    public static void main(String[] args) {
        Scanner rr = new Scanner(System.in);
        int n = rr.nextInt();
        int w=0;
        if ((n>=1)&&(n<=10000)) {
            for(int i=1; i<=n;i++){
                w=w+i;
            }
            System.out.println(w);
        }
        if ((n<=1)&&(n>=-10000)) {
            for (int i=1; i<=Math.abs(n); i++){
                w=w+(-i);
            }
            System.out.println(w+1);
        }
    }
}

Edited by author 13.04.2017 18:52

Why min base number of the sample is 11? In statement I can't find that k should be >= max digit in an initial number.

BTW if I convert any number to binary it will always be divisible by 1 and as far as I see  will satisfy every condition from the statement (the given number, written in k-based system is divisible by k−1, 2 ≤ k ≤ 36).

Thanks.

Edited by author 13.04.2017 17:00

hey guys,this is my solution.Hope you like it :) C++.


#include <bits/stdc++.h>
using namespace std;
queue<int>qu;                           //queue
#define pii pair<int,int>              //pair
#define maxx 500                      //maximum of anything
vector<pii>vpii[maxx];               //Vector of pair
#define inf 0x7fffffff              //INFINITY
int X[]={+2,+2,-2,-2,+1,-1,-1,+1}; //knight movement left/right
int Y[]={-1,+1,-1,+1,+2,+2,-2,-2};//knight movement up/down
int main()
{
    int pr;
    while(cin>>pr){
         int r=max(0,12-pr);
         int minutes=(4*60)-r*45;
         cout<<(minutes>=0?"YES\n":"NO\n");
    }
    return 0;
}

Edited by author 03.06.2015 10:10

Edited by author 03.06.2015 10:10

main {

prepare_lca();
if (N!=-1) {
 build_segement_tree(1,0,N-2); //a[0..n-1]
}
int Q=read();
while(Q-->0) {
 int j=x;
 int x=read()-1;
 int y=read()-1;
 if (x!=y) {
   J= get(1, 0, N-2, x, y-1);
 }
 println(j);
}

}

in build_segement_tree
..
if (tl==tr) t[v]=lca(tl,tl+1);

vertexes are 1..10  and  d , d + 1, d-1, d+2, d-2,  t + 3, t - 3, where d's are divisors of t.

WA 10
WA 14

My Source

import java.util.*;
public class JavaApplication3 {


    public static void main(String[] args) {
        String SubPalabra1="";
        String SubPalabra2="";
        String Correcta="";

        String Verdadera="";
        int menor;
        String resto="";
        Scanner D=new Scanner(System.in);
        String palabra=D.nextLine().trim();
        if(palabra.length()!=1){
            for(int i=0;i<palabra.length();i++){
                Verdadera="";
                resto=palabra.substring(i+1, palabra.length());
                if(resto.indexOf(palabra.charAt(i))!=-1){
                    SubPalabra1=""+palabra.charAt(i)+resto.substring(0,resto.length());
                      SubPalabra2=""+resto.substring(resto.indexOf(palabra.charAt(i)),resto.length());
                    if(SubPalabra1.length()<SubPalabra2.length()){
                        menor=SubPalabra1.length();
                    }
                    else{
                        menor=SubPalabra2.length();
                    }
                    for(int j=0;j<menor;j++){

                        if(SubPalabra1.charAt(j)==SubPalabra2.charAt(j)){
                            Verdadera=Verdadera+SubPalabra1.charAt(j);
                            if(Verdadera.length()>Correcta.length()){
                                Correcta=Verdadera;
                            }
                        }
                        else{
                            j=menor;
                        }
                    }
                }

            }
        }
        else{
            System.out.println(palabra);
        }
        System.out.println(Correcta);


    }
}

Who is the word that no work?

I have written solution that reached test 22 during the competition(I think it corresponds to 15th in offline mode) but now I have wa and i have run out of ideas. Can anyone give me a clue on this problem. Is there a problem with precision (I am making calculations with long double but use sqrt which is a bit unprecise).