In a different IDE with all boundary conditions checked, it's working but it shows run-time error at test #8, not sure why.

import sys
def compute_sum(n):
    return int(n[0]) + int(n[1]) + int(n[2]) - int(n[3]) - int(n[4]) - int(n[5])

n = input()
s2 = 1
s3 = 1
if int(n) == 0 or int(n) == 999999:
    print ('Yes')
    sys.exit()
s1 = compute_sum(n)

if len(str(int(n))) < 6:
    d = 6 - len(str(int(n)))
    s2 = compute_sum('0' * d + str(int(n) - 1))
    s3 = compute_sum('0' * d + str(int(n) + 1))
else:
    s2 = compute_sum(str(int(n) - 1))
    if len(str(int(n) + 1)) <= 6:
        s3 = compute_sum(str(int(n) + 1))

if abs(s2) == 0 or abs(s3) == 0:
    print ("Yes")
else:
    print ("No")

Happy new year, contestants, authors and admins! May this next year bring you happiness, enthusiasm, more AC submits and all that stuff~
Hooray \o/

here's my code.Please give me some advise or test data!

            int n = int.Parse(Console.ReadLine());
            int nCopy = n;
            int sum = 0;
            string[] array = new string[2];
            short digitSum = 0;
            while(n-- != 0)
            {
                digitSum = 0;
                array = Console.ReadLine().Split(' ');
                foreach(var i in array)
                {
                    digitSum += short.Parse(i);
                }
                sum += digitSum * (int)Math.Pow(10, n);
            }

            Console.WriteLine(sum.ToString().Length > nCopy ? sum.ToString().Substring(1):sum.ToString().PadLeft(nCopy,'0'));

can anyone tell me how to solve this question in short memory usage

which is less than o(n*a) or o(n*b)

i will be very thankful to you please help me :D


here is my code

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

int main()
{
    int n,o,t;scanf("%d %d %d",&n,&o,&t);

    int a[500][300]={0};
    int b[500][300]={0};

    int i,j,k;

    for(i=1;i<=n;i++)
    {
        for(j=1;j<=i;j++)
        {
            if(j>o)
            break;

            if(j==i)
            if(j<=o)
            a[i][j]=1; ///which shows it is valid

           a[i][j]=a[i][j]+b[i-j][0];   ///as the places inc with j we will inc in the table of other above
        }

        for(k=1;k<=o;k++)
        a[i][0]+=a[i][k]; ///total sum of all

         for(j=1;j<=i;j++)
        {
            if(j>t)
            break;

            if(j==i)
            if(j<=t)
            b[i][j]=1; ///which shows it is valid

           b[i][j]=b[i][j]+a[i-j][0];   ///as the places inc with j we will inc in the table of other above
        }

         for(k=1;k<=o;k++)
        b[i][0]+=b[i][k]; ///total sum of all

    }

  printf("%d\n",a[n][0]+b[n][0]);


    return 0;
}

Edited by author 13.12.2015 01:13

Edited by author 29.12.2015 22:25

What's on test 6? help, please!

a=int(input())
b=int(input())
print(a+b)

Edited by author 26.12.2015 17:28

1000 => 501501000
500  => 62875500
333  => 18629685
1    => 3
3    => 30
4    => 60
and use long long =)

Sorry for my bad english

I'm see all exists tests, and  I'm get a correct result on these tests(according comments).

Let <ms> is init array, <flag> is (answer is YES?).
1. I do check that (the number in <ms>) is not greater than <n>
2. I'm sorted <ms>.
3. I'm create a <isFills> = new Array[Boolean](n+1)

And then I use the following code:

     if (m>1 && ms(1) == 0) return false  // (ms(0)==ms(1)==0) => "NO"
     if (m>1 && ms(m-2)==n) return false  // (ms(m-2)==ms(m-1)==n) => "NO"

     isFills(max(ms(0)-1, 0)) = true
     for (i <- 1 until m) {
       if (!isFills(ms(i)-1) || !isFills(ms(i))) {
         if (!isFills(ms(i)-1)) isFills(ms(i)-1) = true
         else                   isFills(ms(i))   = true
       } else return false
     }
     return flag

     Console.println(if (getFlag==true) "YES" else "NO")
Full code:
object Main {
   import java.util.Scanner
   import math._

   val scan = new Scanner(System.in)
   val n = scan.nextInt()
   val m = scan.nextInt()
   val ms = new Array[Int](max(m,1))
   for (i <- 0 until m) ms(i) = scan.nextInt()
   scan.close()

   def getFlag: Boolean = {
     var flag = true

     for (i <- 0 until m) flag &&=  ms(i)<=n
     if (!flag) return flag

     java.util.Arrays.sort(ms)
     val isFills = new Array[Boolean](n+1)


     if (m>1 && ms(1) == 0) return false
     if (m>1 && ms(m-2)==n) return false

     isFills(max(ms(0)-1, 0)) = true
     for (i <- 1 until m) {
       if (!isFills(ms(i)-1) || !isFills(ms(i))) {
         if (!isFills(ms(i)-1)) isFills(ms(i)-1) = true
         else                   isFills(ms(i))   = true
       } else return false
     }
     flag
   }
   Console.println(if (getFlag==true) "YES" else "NO")

   def main(args: Array[String]){}
}
Edited by author 26.12.2015 23:56

Edited by author 26.12.2015 23:57

Give me tests!

my program just does dfs three times on the graph.... and so has order O(|n| + |e|) where n is number of nodes and e is edges till gives TLE in case 21... :(
do we need a better algorithm than this as well...?

Can any body tell me whai is in the test case 5...??? My code seems right but everytime it says WA#5...

Edited by author 24.12.2015 20:50

Edited by author 24.12.2015 20:50

Can anybody help me with this test case?

Can anyone help with this?
I do not print at the end of an empty string. all the tests of the discussions my program passes.

maximize amount of teams, but minimize max amount of fighters in a team.
For example, if n=9, k=5, you should make 5 teams this way: (2, 2, 2, 2, 1)
Tests:
Input:
3
7 4
9 4
12 5
Output:
18
30
57

Edited by author 23.12.2015 22:42

I used a little different approach from the one mentioned on this board. It was based on weighted quick-union described by Sedgewick... I made a stupid logical mistake which caused my program to work the longest time possible, so first I got TLE#10 (solutionID=861144). Then I optimized my code with path compression (see Sedgewick once again) and got AC in 0.125 sec (solutionID=861147). Then I realized my error, removed path compression and got AC in 0.078 s (solutionID=861151). Adding path compression back slowed my solution a bit (0.109 s, solutionID=861153). All my solutions consume 449 KB of memory.

I'm interested in time and memory requirements of other possible algorithms. I haven't got their implementations, so I cann't submit them myself.

If somebody feels interested about my approach to this problem, just let me know - I'll explain.

Edit my first comment, see the next one for complete explanation. I can share my code via email if you don't have an access to it.

Edited by author 21.12.2015 22:46

Running in my computer is OK,use use Visual C++ 2010. But in here (use Visual C++ 2013) is wrong answer. I don't understand. Help me, please!

#include <iostream>
using namespace std;

int main(){
        int a[100][100];
        int n, x=0;
        cout<<"\Input n: ";
        cin>>n;
        for(int j=n-1,i=0;j>=0;j--){
            int h=i,k=j;
            while(k<n) a[h++][k++]=++x;
        }
        for(int i=1,j=0;i<n;i++){
            int h=i,k=j;
            while(h<n) a[h++][k++]=++x;
        }
        for(int i=0; i<n ;i++){
            for(int j=0; j<n ;j++){
                cout.width(4);
                cout<<a[i][j];
            }
            cout<<endl;
        }

    cin.ignore(80,'\n');
    cin.get();
    return 0;
}

import java.util.Scanner;
import java.util.Stack;




public class CliperMessage_1654_Stack {
    public static void main(String[] arg1) {
        Scanner sc=new Scanner(System.in);
        StringBuilder sb=new StringBuilder();
        Stack<Character> stc=new Stack<>();
        sb.append(sc.nextLine());
        if(sb.length()<=200000){
            for(int i=0;i<sb.length();i++){
                if(stc.isEmpty()||(stc.peek()!=sb.charAt(i)))
                    stc.push(sb.charAt(i));
                else
                    if((stc.peek()==sb.charAt(i))) stc.pop();

            }

            StringBuilder sbn=new StringBuilder();
            int j=0;
            while(!stc.empty()) sbn.insert(j++, ""+stc.pop());
            System.out.println(sbn.reverse());
        }
    }
}