what can i do if i struck on a problem,becoz there are no solutions n hints availabe here :(

for(int m=i+2;m<=j;m++)
    if((m-i)%2==1)
    {
        t2+=t1;
        if(t2<-8000000000LL||(t2>8000000000LL))//Look at this line.
            break;
    }
    else
    {
        t1+=t2;
        if(t1<-8000000000LL||(t1>8000000000LL))//Look at this line.
            break;
    }

----
    Sorry for my bad English

F1 = 1 F4 = 4
what numbers are F2 and F3?

F2 = F1+F0
F3 = 2*F1+F0
F4 = 3*F1+2*F0

F0 == 0.5???

anzhig@inbox.ru

Edited by author 20.07.2012 18:56

why wrong answer?? help please.
var a,b,c,d:integer;
begin
readln(a,b);
c:=10-a;
d:=10-d;
writeln(c);
writeln(d);
end.

Why always Memory limit exceeded???
It is so wired.
package com.island.timus.ahundrend;

import java.util.Scanner;

public class T1003_Parity {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        int length = in.nextInt();
        while (length != -1) {
            int[][] friendsAndEnemies = new int[2][length + 1];

            for (int j = 0; j < length + 1; j++) {
                friendsAndEnemies[0][j] = j;
                friendsAndEnemies[1][j] = -1;
            }

            int conditions = in.nextInt();
            int counter = 1;
            for (int i = counter; i <= conditions; i++, counter++) {
                Term.start = in.nextInt();
                Term.end = in.nextInt();
                Term.parity = in.nextLine();
                if (Term.start > length || Term.end > length) {
                    break;
                }
                if (Term.parity.trim().equals("even")) {
                    join(friendsAndEnemies[0], Term.start - 1, Term.end);
                } else {
                    int eneymyAncestor1 = findAncestor(friendsAndEnemies[0], Term.start - 1);
                    int eneymyAncestor2 = findAncestor(friendsAndEnemies[0], Term.end);
                    if (friendsAndEnemies[1][Term.start - 1] == -1 && friendsAndEnemies[1][Term.end] == -1) {
                        friendsAndEnemies[1][Term.start - 1] = eneymyAncestor2;
                        friendsAndEnemies[1][Term.end] = eneymyAncestor1;
                    } else if (friendsAndEnemies[1][Term.start - 1] != -1 && friendsAndEnemies[1][Term.end] == -1) {
                        join(friendsAndEnemies[0], friendsAndEnemies[1][Term.start - 1], Term.end);
                        friendsAndEnemies[1][Term.end] = eneymyAncestor1;
                    } else if (friendsAndEnemies[1][Term.start - 1] == -1 && friendsAndEnemies[1][Term.end] != -1) {
                        join(friendsAndEnemies[0], friendsAndEnemies[1][Term.end], Term.start - 1);
                        friendsAndEnemies[1][Term.start - 1] = eneymyAncestor2;
                    } else {
                        join(friendsAndEnemies[0], friendsAndEnemies[1][Term.end], Term.start - 1);
                        join(friendsAndEnemies[0], friendsAndEnemies[1][Term.start - 1], Term.end);
                    }
                }
                int pause = check(friendsAndEnemies);
                if (pause == friendsAndEnemies[0].length) {
                    continue;
                } else {
                    break;
                }
            }

            System.out.println(counter - 1);
            for (int i = counter + 1; i <= conditions; i++) {
                int x = in.nextInt();
                int y = in.nextInt();
                String z = in.nextLine();
            }
            length = in.nextInt();
        }
    }

    private static int findAncestor(int[] friends, int value) {
        int r = value;
        while (r != friends[r]) {
            r = friends[r];
        }
        return r;
    }

    private static void join(int[] pre, int x, int y) {
        int ancestorX = findAncestor(pre, x);
        int ancestorY = findAncestor(pre, y);
        int r = y;
        if (ancestorX != ancestorY) {
            while (r != pre[r]) {
                int temp = pre[r];
                pre[r] = ancestorX;
                r = temp;
            }
            pre[ancestorY] = ancestorX;
        }
    }

    private static int check(int[][] friendsAndEnemies) {
        int i = 1;
        for (; i < friendsAndEnemies[0].length; i++) {
            int friendAncestor = findAncestor(friendsAndEnemies[0], friendsAndEnemies[0][i]);
            int enemyAncestor = -1;
            if (friendsAndEnemies[1][i] != -1) {
                enemyAncestor = findAncestor(friendsAndEnemies[0], friendsAndEnemies[1][i]);
            }
            if (friendAncestor == enemyAncestor) {
                break;
            } else {
                continue;
            }
        }
        return i;
    }

}

class Term {
    public static int start;
    public static int end;
    public static String parity;
}

below is my java code:

import java.util.*;
public class timus1581
{
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        if (t>=1 && t<=1000)
       {
        int i = 0; int cnt = 0;
        int a[] = new int[t];
        for (i = 0; i <= t-1; i++)
        {
            a[i] = sc.nextInt();
          if (a[i]>=1 && a[i]<=10)
          {
              continue;
          }
          else
          {
              System.exit(0);
          }
        }
        for (i = 0; i < t-1; i++)
        {
            if (a[i] == a[i+1])
            {
                cnt++;
            }
            else
            {
                System.out.print((cnt+1) + " " + a[i] + " ");
                cnt = 0;
            }
            if (t == i + 2)
            {
                System.out.print((cnt+1) + " " + a[i]);
                cnt = 0;
            }
        }
      }
    }
}

Edited by author 09.04.2016 14:32

Hi! My code has complexity O(2^N) and it had AC in 0.001 sec. A friend of mine noted this: when I run my program in this case:

50 50
1 2
3 4
5 6
... Same pattern here
95 96
97 98
99 100

And it will give TLE.

Please, add this case!

50 50
1 2
3 4
5 6
7 8
9 10
11 12
13 14
15 16
17 18
19 20
21 22
23 24
25 26
27 28
29 30
31 32
33 34
35 36
37 38
39 40
41 42
43 44
45 46
47 48
49 50
51 52
53 54
55 56
57 58
59 60
61 62
63 64
65 66
67 68
69 70
71 72
73 74
75 76
77 78
79 80
81 82
83 84
85 86
87 88
89 90
91 92
93 94
95 96
97 98
99 100

I have ploted the tour

email: ruben.ashughyan@gmail.com

Edited by author 01.11.2017 22:56

Edited by author 12.05.2021 20:35

Edited by author 01.06.2014 20:04

my code is running correctly but yet it fails at test 1 :(

Can you add Rust language compiler?

Edited by author 11.06.2016 16:54

Just a passing note: the same exact code gets TLE in G++ but AC in Visual C++ (with a very comfortable margin). Make sure to test your code in both compilers before making a judgement that your code is too slow, for those who are coding in C/C++.

Here is my Solution. As this problem is related to stable sort and there is  built in stable_sort function in c++ so it has become easier


#include <bits/stdc++.h>
using namespace std;
bool compare( const pair<long long int,long long int>& x, const pair<long long int, long long int>& y )
{

    return (x.second>y.second);

}
int main()
{
    vector<pair<long long int,long long int> >a;
    long long int n,i,j,b,c;
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>b>>c;
        a.push_back(pair<long long int,long long int>(b,c));
    }
    stable_sort(a.begin(),a.end(),compare); //must include stable_sort
    vector<pair<long long int,long long int> >::iterator p;
    for(p=a.begin();p!=a.end();p++)
    {
        cout<<p->first<<" "<<p->second<<endl;
    }
    return 0;



}

Edited by author 11.06.2016 01:55

Whats wrong with my code. I get correct answer for Test 6 on my compiler, but when I submit the solution, I get wrong answer. Please help!!

I tried with below values and got correct answer
4 3
3 4 5

Correct answer: 1

#include <iostream>
using namespace std;
int main()
{
    int a,k,n,d;
    int b=0;
    cin>>k>>n>>d;
    for (int i=0;i<n-1;i++)
    {
        cin>>a;
        a=a-k;
        b=b+a;
    }
    if (d-k>=0)
    {
        if (b+d-k>=0)
        cout<<b+d-k;
        else cout<<0;
    }


    else if (b>0) cout<<b;
    else cout<<0;
    char f;
    cin>>f;

}

I used Dijkstra+heap optimized but still got TLE on #11..
Is there some better algorithm?
help..
my email:caoyuan@mail.ustc.edu.cn
thx.

What is wrong
#include <cstdio>
unsigned long int contador[65535];
void carga() {
    for(int i=0;i<65535;i++) {
        if(i==0) contador[i] = 1; else contador[i] = contador[i-1] + i; }}
bool busqueda(unsigned long int k[], int i) {
    bool buscador = false;
    int inf = 0, sup = 65534, centro;
    while(inf <= sup) {
        centro = ((sup - inf) / 2) + inf;
        if(contador[centro] == k[i]) {
            buscador = true;
            break; }
        if(contador[centro] < k[i]) { inf = centro + 1; }
        if(contador[centro] > k[i]) { sup = centro - 1; }}
    return buscador; }
int main() {
    carga();
    unsigned long int k[65535];
    int n, i;
    scanf("%d", &n);
    for(int i=0;i<n;i++) {
        scanf("%lud", &k[i]); }
    for(i=0;i<n-1;i++) {
        if(busqueda(k, i)) printf("1 "); else printf("0 "); }
    if(busqueda(k, i)) printf("1"); else printf("0");
    printf("\n"); }

It was very easy.
Backtracking(перебор с возвратом)

.

Edited by author 13.06.2016 15:16

wrong answer 3