The delete operation is used at most once when finding a name!

As I have encountered, I think the test is most likely several same length sequences which are not contained by any. Make sure you try this case before submitting :)

Guys,
Can somebody give me a hint to avoid Memory Limit (I use C#)?

I've tried a lot of algorithms:
1. Composing full graph by using digits tree for all numbers. Processing graph by using Dijkstra and PriorityQueue (SortedArray<ShortestDistance, Queue<VertexIndex>>).
2. Creating tree of digits for all numbers. Processing graph "on the fly": obtaining all neighbor vertexes for current vertex (from previously created tree). The Dijkstra and the same PriorityQueue were used.
3. Using Dictionary<PhoneNumber, VertexIndex> for storing all numbers and vertexes. Processing graph "on the fly": obtaining all possible neighbors (135) and filter them by real vertexes from dictionary. The same PriorityQueue was used.
4. The same as 3, but instead of PriorityQueue, I've used Heap<ShortestDistance, VertexIndex>. After distance to some vertex was updated, I remove this vertex from Heap (if it exists there) and add it again, but with new key (distance).

I even replaced usage of all pointers to objects by UShort indexes, and now I'm in stuck.
I've started work on this task because accidentally send solution of other task here, but now I already hate test #8 :) Any ideas?

solved :x

Edited by author 26.03.2012 03:51

import java.util.Scanner;
public class Bil {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int a = input.nextInt();
        int b = numbers (Math.abs(a));
        if (a >= 0 )System.out.println(b);
        else {System.out.println(-b);}
    }


    public static int numbers( int a){
        int sum = 0;
        if (a > 1){
            for (int i = 2; i <= a; i++)
            sum +=i;
            }
//        else if (a == 0) sum = 0;
        else{
            sum = 1;
            }
        return sum;
    }
}

// Nvidia CUDA Driver API 4.1
//========================
#include <cuda.h>
#include <stdio.h>
#include <device_launch_parameters.h>
#include <Windows.h>

//========================
#pragma comment(lib,"cuda.lib")

//========================
int wmain()
{
    //
    CUdevice     cuDev;
    CUcontext    cuCtx;
    CUmodule     cuModl;
    CUfunction   cuFunc;
    //
    CUdeviceptr  pDmem[3];
    CUstream     stream[4];
    CUevent      start;
    CUevent      end;
    size_t       paramsize;
    void*        kernelparams[4];
    float        time;
    //
    cuInit(0);
    cuDeviceGet(&cuDev,0);
    cuCtxCreate(&cuCtx,CU_CTX_SCHED_AUTO,cuDev);
    //
    cuEventCreate(&start,0);
    cuEventCreate(&end,0);
    //
    for(int i=0; i<4; i++)
    {
        if(cuStreamCreate(&stream[i],0))
        {
            return 0;
        }
    }
    //
    cuModuleLoad(&cuModl,"matrixtile.ptx");
    cuModuleGetFunction(&cuFunc,cuModl,"add");
    //
    for(int i=0; i<3; i++)
    {
        if(cuMemAlloc(&pDmem[i],sizeof(int)))
        {
            return 0;
        }
    }

    int a = 1;
    int b = 8;
    int r;
    cuEventRecord(start,0);
    if(cuMemcpyHtoDAsync(pDmem[0],(void*)&a,sizeof(int),stream[0]))
    {
        return 0;
    }
    if(cuMemcpyHtoDAsync(pDmem[1],(void*)&b,sizeof(int),stream[1]))
    {
        return 0;
    }
    if(cuMemcpyHtoDAsync(pDmem[2],(void*)&r,sizeof(int),stream[2]))
    {
        return 0;
    }
    //
    paramsize = sizeof(pDmem);
    kernelparams[0] = &pDmem[0];
    kernelparams[1] = &pDmem[1];
    kernelparams[2] = &pDmem[2];
    kernelparams[3] = &paramsize;

    //
    cuLaunchKernel(cuFunc,1,1,1,1,1,1,0,stream[3],kernelparams,0);
    cuEventRecord(end,0);
    cuEventSynchronize(end);
    cuEventElapsedTime(&time,start,end);
    wprintf(L"Time = %.7f\n\n",time);
    //
    cuMemcpyDtoHAsync(&r,pDmem[2],sizeof(int),stream[3]);
    cuCtxSynchronize();
    //
    cuModuleUnload(cuModl);
    cuCtxDestroy(cuCtx);
    wprintf(L"result = %d\n",r);
    //
    ReadFile(GetStdHandle(STD_INPUT_HANDLE),0,0,0,0);
    return 0;
}

Time = 0.0076 milliseconds

Edited by author 25.03.2012 21:23

#include<iostream>
#include<stdio.h>
#include<string>
using namespace std;
int main() {
    int desk[][4] = { { 2, 3, 4, 4 }, { 3, 4, 6, 6 }, { 4, 6, 8, 8 }, { 4, 6,
            8, 8 } };
    int n;
    scanf("%d", &n);
    string str;
    getline(cin, str);
    char c, c1;
    int x, y;
    for (int i = 0; i < n; i++) {
        scanf("%c%c", &c, &c1);
        x = (c - 'a') > 3 ? 7 - (c - 'a') : (c - 'a');
        y = (c1 - '1') > 3 ? 7 - (c1 - '1') : (c1 - '1');
        printf("%d\n", desk[x][y]);
        getline(cin, str);
    }
    return 0;
}

Does the checking system enter only prime number ('p')? Should I do a check of input data?

And

What is the distinction of test 8? I broke  my head, trying to solve the WA!

Edited by author 25.03.2012 18:41

1) What does it mean by two-way road?

Lets say I use road 1 and 3 in the sample..
To go from 1 to 3, it costs 300 and from 3 to 1 it is just 10. In that case we dont have route to go from 5 to 3. Then why is the answer correct?

1 3 5 2

1 to 3 -> 300
3 to 5 -> (No route, but taking 5 to 3 value) 300 + 20
5 to 2 -> (Again no route, taking 2 to 5 value) 300 + 20 + 15
2 to 1 -> (No route, using 1 to 2) 300 + 20 + 15 + 16 => 351

But consider 1 2 5 3

1 to 2 -> 16
2 to 5 -> 16 + 15
5 to 3 -> 16 + 15 + 20
3 to 1 -> 16 + 15 + 20 + 10 which is just 61

Could someone please explain why it is 1 3 5 2 and not 1 2 5 3?

WA #39
Did somebody have the same problem?

Edited by author 24.03.2012 14:17

Edited by author 24.03.2012 14:17

1.The relationship is always mutual,which means that if 1 has a friend 2,then 2 has a friend 1.
2.1 has a friend 2, and 2 has a friend 3 doesn't mean 1 and 3 are friends,thus be careful to use disjoint set.

Hope they are helpful to you guys :)

In C++ don't use cout and cin! Just use scanf(...) and printf(...).
Enjoy!

Edited by author 24.03.2012 00:22

Big number really?

In http://timustests.lx.ro/, the test solution 8 is "****** bargo ********" but the word "bargo" is not in the dictionary.

help.
send please the test 1 for mail antonluzhbin@yandex.ru

Does it contains only ['a'-'z'] and ['A'-'Z']?

actually i dont have any idea for this my sir gave me this assessment i want to solve this problem please help me how to start the program and which concepts we want to use

When n = 0,
if we have to calculate T, is the answer 'undefined' ?
if we have to calculate V, is the answer 'error' or 'undefined'?