My programm search the greatest angle in x in [x1..x2] and got AC.
But it's wrong. For example in test 0 3 1 1.

PS Maybe i'm wrong

Edited by author 13.01.2011 21:20

program correctly solves all problems, but for some reason the site shows that the program is wrong ... can you explain why?

#include<iostream>
using namespace std;

int main()
{
    double ves_k[20];
    int i,num;
    double t,perv,vtor,res;

    cin>>num;
    if(num<1||num>20) goto vozvrat;
    for(i=0; i<num; i++){
        cin>>ves_k[i];
        if(ves_k[i]<1||ves_k[i]>10000) goto vozvrat;
    }
    if(num!=1){
    for(int a=1; a<i; a++)
        for(int b=i-1; b>=a; b--){
            if(ves_k[b-1]>ves_k[b]) {
                t=ves_k[b-1];
                ves_k[b-1]=ves_k[b];
                ves_k[b]=t;
            }
        }
    perv=ves_k[num-1];
    vtor=ves_k[num-2];
    for(i=num-3; i>=0; i--){
        if(perv>=vtor) vtor+=ves_k[i];
        else perv+=ves_k[i];
    }
    res=perv-vtor;
    if(res<0) res=-res;
    } else res=ves_k[0];

    cout<<res;
vozvrat:
    return 0;
}

Edited by author 13.01.2011 02:22

Good time of day.
I constantly get WA on test #6. Does anybody know why may this happen?
Here's my code, seems to give correct answers for any correct test found in discussions:

#include <stdio.h>
#include <iostream>
#include <map>

using namespace std;

typedef unsigned int uint;

const uint MAX_N = 5000;
const bool COLOR_BLACK = false;
const bool COLOR_WHITE = true;

typedef pair <uint, bool> LinePair;
typedef map<uint, LinePair> TLines;

TLines lines;

void paint(uint c, uint d, const bool clr)
{
    if (c == d)
        return;
    uint a, b;
    bool c2;
    if (lines.empty())
    {
        lines[c] = LinePair(d, clr);
        return;
    }

    TLines::iterator i = lines.begin();
    //пропускаем все отрезки, что левее нашего и не накладываются на него
    while (c > (i->second).first) ++i;
    //работаем с накладывающимися отрезками
    while ((d >= i->first) && (i != lines.end()))
    {
        //для удобства написания
        a = i->first;
        b = (i->second).first;
        c2 = (i->second).second;
        if (a >= c)
        {
            if (d >= b)
            {
                lines.erase(i);
            }
            else if (clr == c2)
            {
                d = b;
                lines.erase(i);
            }
            else
            {
                lines.erase(i);
                lines[d] = LinePair(b, c2);
            }
        }
        else
        {
            if (d >= b)
            {
                if (clr == c2)
                {
                    c = a;
                    lines.erase(i);
                }
                else
                {
                    (i->second).first = c;
                }
            }
            else if (clr == c2)
            {
                c = a;
                d = b;
            }
            else
            {
                (i->second).first = c;
                lines[d] = LinePair(b, c2);
            }
        }
        ++i;
    }
    //собственно добавление заданного отрезка
    lines[c] = LinePair(d, clr);
    return;
}

int main()
{
    uint n, b, e;
    char ch;
    bool clr;
#ifndef ONLINE_JUDGE
    freopen("input.txt", "rt", stdin);
    //freopen("output.txt", "wt", stdout);
#endif
    lines.clear();
    paint(0, 1e+9, COLOR_WHITE);
    cin >> n;
    for (uint i = 0; i < n; i++)
    {
        cin >> b >> e >> ch;
        if (ch == 'b')
            clr = COLOR_BLACK;
        else
            clr = COLOR_WHITE;
        paint(b, e, clr);
    }
    TLines::const_iterator maxLine;
    uint max = 0;
    for (TLines::const_iterator i = lines.begin(); i != lines.end(); ++i)
    {
        if (((i->second).second == COLOR_WHITE) && ((i->second).first - i->first > max))
        {
            max = (i->second).first - i->first;
            maxLine = i;
        }
#ifndef ONLINE_JUDGE
        cout << i->first << '\t' << (i->second).first << '\t' << (i->second).second << endl;
#endif
    }
    cout << maxLine->first << " " << (maxLine->second).first << endl;
    return 0;
}

Edited by author 05.01.2011 03:29

First, very stupied method: used just 2 arrays for A and B values, 0.609s, 673KB.
Second, some hint: used multi-stack and for pointers at next element used unsigned short, 0.031s, 677KB.
Sorry, for bad english:-)

Hello,
I try to solve problem 1209 with allocating memory, but i receive compilation error.

---------------------------------------------------------
e2c33114-a1cd-488e-b50b-9a2950afaee7
e2c33114-a1cd-488e-b50b-9a2950afaee7(8) : error C2143: syntax error : missing ';' before 'type'
e2c33114-a1cd-488e-b50b-9a2950afaee7(19) : error C2065: 'AA' : undeclared identifier
e2c33114-a1cd-488e-b50b-9a2950afaee7(19) : error C2109: subscript requires array or pointer type
e2c33114-a1cd-488e-b50b-9a2950afaee7(24) : error C2065: 'AA' : undeclared identifier
e2c33114-a1cd-488e-b50b-9a2950afaee7(24) : error C2109: subscript requires array or pointer type
e2c33114-a1cd-488e-b50b-9a2950afaee7(30) : error C2065: 'AA' : undeclared identifier
e2c33114-a1cd-488e-b50b-9a2950afaee7(30) : error C2109: subscript requires array or pointer type
e2c33114-a1cd-488e-b50b-9a2950afaee7(32) : error C2065: 'AA' : undeclared identifier
e2c33114-a1cd-488e-b50b-9a2950afaee7(32) : warning C4022: 'free' : pointer mismatch for actual parameter 1
---------------------------------------------------------


Code example:
---------------------------------------------------------
#include <stdio.h>
#include <malloc.h>

int main()
{
        int N;
    scanf("%d", &N);
    int *AA = (int *)malloc(N*sizeof(int));
        ... ... ...
        free(AA);
        return 0;
}
---------------------------------------------------------
What the matter ?

Thank you for answers.

Edited by author 12.01.2011 23:44

You can go from last digit to the first (the leftmost digit to the rightmost digit).

For example : 302

Considering 2 : There are some possibilities: 0 + 2, 1 + 1, 2 + 0, 6 + 6, 5 + 7 ... (first operant is of the larger number, second operant is of the smaller number).

Considering 0 : For each possibilities, you try to find out some possibilities that can be combined to produce 0 : 0 + 0...
You can try any combinations to produce 0 if two operants used to produce 2 is equal, if they are different, you only have an only way, it reduce very much calculations).

Considering 3 : The last digit, you can stop here and consider whether you have achieved a solution.

It helped me to get AC.

Test:
1 +
1 +
1 +
10 . 2
11 . 3
600 +
601 . 1
609 . 2
611 . 3

Right answer:
1
2
3
+
+
4
-
+
-

Hello!

Is it allowed to use multiple threads in solutions?

Thank you.

Edited by author 10.01.2006 12:58

Please, give me tests or hints about test 1!!!

Edited by author 11.01.2011 16:16

I enjoyed solving that problem! Very interesting, thanks a lot.

P.S. Answer for empty grid 12x12 = 1076226888605605706

My O(M*M*log(N)) solution passed.

Can you show us Input with the Wrong Answer result?

Any program written on C# gets OLE (unless it gets CE).

Can explain please that tests - 1, 2, 3, 4, 5 mean.То there is what childbirth of errors they reveals.

5 6
1 5 1
3 5 100
3 4 2
2 4 1
4 5 10
1 2 1

Answer:
1 2 4 5

50 authors lost AC after rejudge.

I tried to solve with some float precisions but it was always WA. Then it was written using only integer calculations and AC.

So, if the pigeon is black or white, how does this work?

Edited by author 06.01.2011 15:52