I had WA7....corrected one error but now getting WA18...Mods please give me 18th test case...
Thanks...

Edited by author 09.01.2009 13:12

Some of my AC solutions don't pass this test:
3
9999 10000 1 A
-9990 9999 999 B
-10000 9999 1000 C
1
9999 10000 -10000 9998

Right answer:
Power on. CELL_ID:A, SIGNAL_LEVEL:VIOLET
Signal changed. SIGNAL_LEVEL:INDIGO
Signal changed. SIGNAL_LEVEL:BLUE
Signal changed. SIGNAL_LEVEL:GREEN
Signal changed. SIGNAL_LEVEL:YELLOW
Signal changed. SIGNAL_LEVEL:ORANGE
Signal changed. SIGNAL_LEVEL:RED
Cell changed. CELL_ID:C, SIGNAL_LEVEL:ORANGE
Signal changed. SIGNAL_LEVEL:YELLOW
Signal changed. SIGNAL_LEVEL:GREEN
Signal changed. SIGNAL_LEVEL:BLUE
Signal changed. SIGNAL_LEVEL:INDIGO
Signal changed. SIGNAL_LEVEL:VIOLET

Edited by author 05.11.2014 21:55

Edited by author 05.11.2014 21:56

input:
2
1 600000
1
600001

output:
599999

The problem is not that difficult, therefore for me the most interesting question is the smallest program possible to write for this problem.
Let's define the size of the program as H x W in your solution. Mine is 6 x 8. Who's smaller? =)

For big tests (at least for n >= 44, but may be works for smaller n's) you can assume that firstt 4 numbers are 33 11 22 44

Edited by author 07.10.2024 02:40

Why I have WA#10? I can't think test to get wrong answer. PLS help me

Next my code:

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

int main()
{
    double a, b, c, x, y, x2, y2;
    double X, Y, Z, X2, Y2, Z2;
    double result = 0;
    bool bflag = true;
    cin >> a >> b >> c;
    for (int i=0; i < 2; i++)
    {
        cin >> x >> y;
        if (x < c)
        {
            X = 0;
            Y = y - b - c;
            Z = x;
        }
        else if (x > c + a)
        {
            X = a;
            Y = y - b - c;
            Z = 2 * c + a - x;
        }
        else
        {
            if (y <= b)
            {
                X = x - c;
                Y = b - y;
                Z = 0;
            }
            else if (y <= b + c)
            {
                X = x - c;
                Y = 0;
                Z = y - b;
            }
            else if (y <= b)
            {
                X = x - c;
                Y = y - b - c;
                Z = c;
            }
            else
            {
                X = x - c;
                Y = b;
                Z = 2 * b + 2 * c - y;
            }
        }
        if (bflag)
        {
            x2 = x;
            y2 = y;
            X2 = X;
            Y2 = Y;
            Z2 = Z;
            bflag = false;
        }
    }
    //cout << X << ' ' << Y << ' ' << Z << ' ' << X2 << ' ' << Y2 << ' ' << Z2; это типа отладка
    X2 -= X;
    Y2 -= Y;
    Z2 -= Z;
    result = sqrt(X2 * X2 + Y2 * Y2 + Z2 * Z2);
    if (result < 1.E-8)
        cout << fixed << setprecision(6) << 0;
    else
        cout << fixed << setprecision(6) << result;

    return 0;
}

подскажите в какие там данные входные

If I delete a branch, does it means that I delete also its sub-braches?

#include <cstdio>

int ncall;
int call[100];

void solve(int K)
{
    if (!K) return;

    if (K % 2 == 0) {
        call[ncall] = ++ncall;
        if (K > 2) solve(K / 2);
    } else {
        call[ncall++] = -1;
        solve(K - 1);
    }
}

int main( void )
{
//    freopen( "p1278.in", "r", stdin );

    int K;
    scanf( "%d", &K );

    if (K > 1) solve(K);
    for (int i = 0; i < ncall; i++)
        printf( "CALL %d\n", call[i] < 0 ? ncall : call[i] );
    printf( "BELL&RET\n" );

    return 0;
}

When finding intersection of segments (a, b) and (c, d) I am finding t1, t2, such that a + (b - a) * t1 = c + (d - a) * t2, -eps <= t1 <= 1+eps, -eps <= t2 <= 1+eps.
This code doesn't work for eps=1e-12, but works for eps=0.

THIS IS MINDBLOWING

Do you know the test#3?
i use heap in this problem but got WA#3.
please help.
sorry for poor english.
thank.

Use stacks.
Push one letter by letter.
If stack.top() == current letter, pop the stack.
else push the letter in the stack.

wrong epsilon, precision problems

def get_max(arr):
    res = arr[0]
    maxEnding = arr[0]
    for i in range(1, len(arr)):
        maxEnding = max(maxEnding + arr[i], arr[i])
        res = max(res, maxEnding)

    return res

Some hints for solving this problem the way I was able to solve it, without greedy search

1. I used BFS search with heuristic (A* search). As a heuristic, I used the sum of the distances from the boxes to the nearest goals, taking into account that different boxes should be on different goals. It is better to distribute boxes among goals in some greedy way, so as not to spend a lot of time on this.
2. In the board states, DO NOT store the position of the player, store the places he can REACH. This will greatly reduce the number of states needed to be stored. You can store one board state in two 64-bit numbers as bit masks.
3. To find bad positions and stop further search on them, look for simple deadlocks (the box is in a corner and not on a target), dynamic deadlocks (the boxes block each other) and NOT COMPLEX corral deadlocks (the boxes are not on goals and block access to some board area, so they cannot be pushed out of there).
4. If you try to search for too complex corral deadlocks, the time spent on finding them will not be worth it. Try different settings of what difficulty of corral deadlocks to search for and when to stop the search. Store the found configurations of corral deadlocks (and the configurations in which no deadlock was found) in some sets so as not to determine them every time.
5. In addition to the forward search (pushing boxes from the starting positions to the goals), you can also use the backward search (pulling boxes from the goals to the starting positions) to reduce the depth of the search tree.
6. For the forward (backward) search, prevent situations when, for example, there are more (fewer) boxes near the wall than goals. This can greatly speed up the search for some boards.
7. To understand why your algorithm is working too long for a particular board, you can find and output long deadlock branches - a sequence of pushes/pulls in the search tree that starts from one of the board states that is in the found solution and that does not eventually lead to the solution. This way you can find bugs when the algorithm did not find one of the types of deadlocks and did not stop searching on them.

Good luck!

если второй пересыпает первому, то у первого должно быть больше, а не наоборот!

А если это a1 и b1, соответственно, то это и есть ответ! Потому что a1 и b1 это изначальные мерки, до пересыпаний, в задаче так описано

Edited by author 22.09.2024 01:35

Edited by author 22.09.2024 01:36

Edited by author 22.09.2024 01:36

Edited by author 22.09.2024 01:37

What could be the reason???

I think it sholud be said in statement that we do rounding to near number. I spent 20 minutes, understanding why my code doesn't work for sample test, but the mistake was here: floor(100*x)/100, because I thought we should round to bottom

Who can help me?
thanks.
Give me some tests.

Здравствуйте!
Не воспринимает компилятор директивы
#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
#include <tuple>
#include <string>

Ошибка:
fatal error: map: No such file or directory
    2 | #include <map>
      |          ^~~~~
compilation terminated.

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