I solve the system by the Gauss method, but the 16-th test he passes.
Then I tried to take first the Gauss method and the results of the initial matrix method of Seidel (iterative algorithm), but it does not always converge and the result becomes invalid even before the 16-second test.
What is there a problem?

I've tried the heap one, I've done everything I could, but still TLE. I guess there should be some tricky part or my implementation error.

what have you guys done to avoid the TLE?

I use heap to arrange the time, everytime there is a new timestamp that is different from the last one, I check the heap and delete the out-of-time nodes.

I also use a bool array to record the existance of a node, as well as a link table to record the next label that can be used. I don't know why I got TLE all the time...

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

char s[10001],str[80];
int main()
{

#ifndef ONLINE_JUDGE
   freopen("input.txt", "rt", stdin);
   freopen("output.txt", "wt", stdout);
#endif

         int j = 0;
         memset(s,'\0',sizeof(s));
        while(gets(s))
        {
                int len = strlen(s);
                int i = 0;
                while(i<len)
                {
                        if(s[i]=='>')
                        {
                                j++;
                                if(j>79)j=0;
                        }
                        else if(s[i]=='<')
                        {
                                j--;
                                if(j<=0)j=0;
                        }
                        else if(s[i]!='\n' && s[i]!='\0')
                        {
                                str[j] = s[i];
                                j++;
                        }
                        i++;
                }

                memset(s,'\0',sizeof(s));
        }
for(int i = 0;i<80;i++)
printf("%c",str[i]);

return 0;
}

New tests have been added. 91 authors have lost AC after rejudge. Thanks to Sergey Kopeliovich.

you should find nearest prime number to J if the number doesnt exist you should find odd number because even number's divisor sum is bigger:)

Edited by author 30.05.2010 18:53

Edited by author 30.05.2010 18:53

Edited by author 30.05.2010 18:54

var
a,c,b : array[1..10000]of longint;
var n : longint;

procedure sort(m,t:longint);
var y,i,j,w:longint;
begin
    i:=m;j:=t;y:=a[(m+t)div 2];
    repeat
         while a[i]<y do inc(i);
         while a[j]>y do dec(j);
         if i<=j then begin
             w:=a[i];a[i]:=a[j];a[j]:=w;
             inc(i);dec(j);
         end;
    until i>j;
    if i<t then sort(i,t);
    if m<j then sort(m,j);
end;

PRocedure red;
var
i,j : longint;
begin
  readln(n);
    for i:=1  to n do
       read(a[i]);

sort(1,n);
end;

{=-=================}
PRocedure get;
var
i,sum1,sum2 : longint;
begin
sum1:=a[n];
sum2:=0;
   for i:=n-1  downto 1 do
   begin
      if sum1>=sum2 then
         begin
            sum2:=sum2+a[i];
         end
         else
         begin
           sum1:=sum1+a[i];
         end;
   end;
   writeln(abs(sum1-sum2));
end;

{=-=================}
BEgin
    red;
    get;
end.

import java.util.Scanner;

public class Queue {

    public static void main(String[] args) {

        Scanner input = new Scanner(System.in);
        koor A[] = new koor[50];
        int i, j, st;
        int vaxt = 0;
        int temp, temp1, temp2;
        int max = 0;
        int N = input.nextInt();

        for (i = 1; i <= N; i++) {
            A[i] = new koor();
            A[i].T1 = input.nextInt();
            A[i].T2 = input.nextInt();
            A[i].T3 = input.nextInt();
        }


        for (i = 1; i <= N - 1; i++) {

            for (j = 1; j <= N - 1; j++) {

                if (A[j].T1 > A[j + 1].T1) {

                    temp = A[j].T1;
                    A[j].T1 = A[j + 1].T1;
                    A[j + 1].T1 = temp;

                    temp1 = A[j].T2;
                    A[j].T2 = A[j + 1].T2;
                    A[j + 1].T2 = temp1;

                    temp2 = A[j].T3;
                    A[j].T3 = A[j + 1].T3;
                    A[j + 1].T3 = temp2;

                }
            }
        }


        st = A[1].T1 + A[1].T2;

        if (st > A[1].T3) {
            vaxt = st - A[1].T3;
        }
        for (i = 2; i <= N; i++) {

            if (st > A[i].T1) {
                st += A[i].T2;
                if (st > A[i].T3) {
                    vaxt = st - A[i].T3;
                }
            } else {
                st = A[i].T1 + A[i].T2;
                if (st > A[i].T3) {
                    vaxt = st - A[i].T3;
                }
            }
            if (vaxt > max) {
                max = vaxt;
            }

        }
        System.out.println(max);
    }
}

class koor {

    int T1;
    int T2;
    int T3;
}

Edited by author 30.05.2010 15:45

help, please

The only input line contains two integers separated with a space: N and K (0 < N ≤ 109, 1 < K ≤ 109).
and There's a Test case with K = 1 which made me got 2 wrong answer...So funny

could anyone tell me,what is test 1 in 1014 ?

4 6
1 2 2
2 3 2
3 4 2
4 1 2
1 3 1
2 4 1
Cost: 4
Cost: 4
1:
why not 1>4(cost 2, minimum. Description says if transition between A > B is possible then B > A is possible too)
That gives answer: 2
2:1>3(cost 1) then 3>4(cost 2), 1+2=3
That gives answer: 3

Anyone mind explaining why its 4 and 4 in sample?

Anybody knows?

Edited by author 25.05.2010 20:52

Please explain me this problem>Thank!!!

What I did was of course order 3 but it got AC in 0.062sec and 1133 kb memory. I wonder why n^3 would need .8 sec or TLE. I may do it in n^2 but please tell me about the infamous O(n) algo.

What is the test 25?

I've already solved the problem, but I want know why my first solution is wrong...

Anybody use this? It's right formula or no?

a1:=(n*a(0)+a(n+1)-sum(i=1 to n+1 , c(i)*2^i))/(n+1)   ?

Edited by author 31.10.2009 19:00

a1 и a2 - соседи? (у них есть общая вершина)

If no, please, tell me, where can I find class BigInteger. Such one, that I wrote myself, works too slow :(

#include <iostream>
#include <algorithm>
#include <string>
#include <cctype>
#include <iomanip>
#include <cmath>
using namespace std;
#define MaxN 500000
#define f0(i,n) for(i=0;i<n;i++)
#define f1(i,n) for(i=1;i<n;i++)
char x[MaxN];
int main()
{
    int i=0;
    char c;
    while((c=getchar())!='\n')
    {
        x[i]=c;
        i++;
    }
    long int total=0,t=1;
    int j,k,l,m;
    m=i;
    int count=0;
    int counter;
    f1(i,MaxN)
    {
        if(i<10 && i>=0)t=1;
        if(i<100 && i>=10)t=2;
        if(i<1000 && i>=100)t=3;
        if(i<10000 && i>=1000)t=4;
        if(i<100000 && i>=10000)t=5;
        f0(j,m)
        {
            total=0;
            counter=1;
            for(k=j;k<j+t;k++)
            {
                l=x[k]-48;
                total+=pow(10,t-counter)*l;
                counter++;
            }
            if(total==i)
            {
                count++;
                break;
            }
        }
        if(count==0)
        {
            cout<<i<<endl;
            break;
        }
        count=0;
    }
    return 0;
}
Thanks!!!