In case anybody wants to see implementation Link is here http://paste.ubuntu.com/13082792/.

Could anyone help me?

Написал лобовой вариант и всё равно WA10! Значит, вывод не верен. Кто может подсказать, как выводить ответ?(Форум не помог)

#define _CRT_SECURE_NO_WARNINGS
#define mk make_pair

#include <string>
#include <map>
#include <iostream>
using namespace std;

typedef map <string, int> mymap;

const int nmax = 100005;

mymap t[4 * nmax];
mymap answer[nmax];
string word[nmax];
int cnt[nmax];
int n, m;
string w_print[11];
int c_print[11];

mymap Search(int L, int R);
int BinaryDown(int L, int R, string s);
int BinaryUp(int L, int R, string s);
mymap Optimal(mymap map1, mymap map2);
void Print(mymap M);
void QSort_W(int L, int R);
void QSort2(int L, int R);
void QSort3(int L, int R);

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> word[i] >> cnt[i];
    QSort_W(1, n);
    cin >> m;
    for (int i = 1; i <= m; i++)
    {
        string s;
        cin >> s;
        int L, R;
        L = BinaryDown(1, n, s);
        R = BinaryUp(1, n, s);
        answer[i].clear();
        if (L != -1 && R != -1) answer[i] = Search(L, R);
    }

    for (int i = 1; i <= m; i++)
    {
        Print(answer[i]);
        if(i!=m) cout<<endl;
        if(!(answer[i].empty()) && i!=m) cout<<endl;
    }

    return 0;
}

mymap Search(int L, int R)
{
    mymap res;
    res.clear();
    for(int i=L;i<=R;i++)
    {
        mymap x;
        x.insert(mk(word[i],cnt[i]));
        res=Optimal(res,x);
    }
    return res;
}

void Print(mymap M)
{
    int n = 0;
    for (mymap::iterator it = M.begin(); it != M.end(); it++)
    {
        w_print[++n] = it->first;
        c_print[n] = it->second;
    }
    if (n == 0) return;
    QSort2(1, n);
    for (int i = 1; i <= n;)
    {
        int j = i + 1;
        while (j <= n && c_print[j] == c_print[i]) j++;
        QSort3(i, j - 1);
        i = j;
    }
    for (int i = 1; i<n; i++) cout << w_print[i] << endl;
    cout << w_print[n];
}

void QSort2(int L, int R)
{
    int i, j;
    int X = c_print[(L + R) / 2];
    i = L, j = R;
    while (i <= j)
    {
        while (c_print[i] > X) i++;
        while (c_print[j] < X) j--;
        if (i <= j)
        {
            string Y = w_print[i]; w_print[i] = w_print[j]; w_print[j] = Y;
            int k = c_print[i]; c_print[i] = c_print[j]; c_print[j] = k;
            i++;
            j--;
        }
    }
    if (L < j) QSort2(L, j);
    if (i < R) QSort2(i, R);
}

void QSort3(int L, int R)
{
    int i, j;
    string X = w_print[(L + R) / 2];
    i = L, j = R;
    while (i <= j)
    {
        while (w_print[i] < X) i++;
        while (w_print[j] > X) j--;
        if (i <= j)
        {
            string Y = w_print[i]; w_print[i] = w_print[j]; w_print[j] = Y;
            int k = c_print[i]; c_print[i] = c_print[j]; c_print[j] = k;
            i++;
            j--;
        }
    }
    if (L < j) QSort3(L, j);
    if (i < R) QSort3(i, R);
}


void QSort_W(int L, int R)
{
    int i, j;
    string X = word[(L + R) / 2];
    i = L, j = R;
    while (i <= j)
    {
        while (word[i] < X) i++;
        while (word[j] > X) j--;
        if (i <= j)
        {
            string Y = word[i]; word[i] = word[j]; word[j] = Y;
            int k = cnt[i]; cnt[i] = cnt[j]; cnt[j] = k;
            i++;
            j--;
        }
    }
    if (L < j) QSort_W(L, j);
    if (i < R) QSort_W(i, R);
}

int BinaryDown(int L, int R, string s)
{
    for(int i=L;i<=R;i++) if(s==word[i].substr(0, s.length())) return i;
    return -1;
}

int BinaryUp(int L, int R, string s)
{
    for(int i=R;i>=L;i--) if(s==word[i].substr(0, s.length())) return i;
    return -1;
}


mymap Optimal(mymap map1, mymap map2)
{
    mymap res;
    res.clear();
    if (map1.size() + map2.size() <= 10)
    {
        res = map1;
        for (mymap::iterator it = map2.begin(); it != map2.end(); it++) res.insert(mk(it->first,it->second));
        return res;
    }
    else
    {
        res = map1;
        for (mymap::iterator it = map2.begin(); it != map2.end(); it++)
        {
            if (res.size() < 10) res.insert(mk(it->first, it->second));
            else
            {
                if (res.begin()->second < it->second)
                {
                    res.erase(res.begin());
                    res.insert(mk(it->first, it->second));
                }
            }
        }
        return res;
    }
}

My AC code:

input:
>   3 3
>   1 2 1
>   1 2 1
>   1 3 1
>   -1

output:
>   No solution.

Why not
>   1 2

There are two roads connecting "1" and "2", so the route could be 1 - 2 - 1, with different road 1 - 2 and 2 -1

Have you guys thought about exact, long arithmetic check?

if you get it, try to change g. i misread the statement and my g was equal to 9.81, not 9.8, then i fixed it and got AC

Maybe correct name for that problem is "SINE dance"?
"Sinus" is something about anathomy.

What is in #1???

What's should the following input get?

3 4
1 2 1
2 3 1
1 3 1
1 2 1

3 4
1 2 1
2 3 1
1 3 1
1 2 10

-1

Thanks!

RT

Precision exactly 4 digits (not >= 4).

38 41
???

2
0 0
0 2
2
2 0
2 2
1
1 -1

->

2 (not 1)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {

            int count = 0, j, n, m, t, i;
            n = int.Parse(Console.ReadLine());
            int[] p_dat = new int[n];

            for (i = 0; i < n; i++)
            {
                p_dat[i] = int.Parse(Console.ReadLine());

            }
            m = int.Parse(Console.ReadLine());
            int[] s_dat = new int[m];
            for (i = 0; i < m; i++)
            {
                s_dat[i] = int.Parse(Console.ReadLine());
            }
            for (i = 0; i < n - 1; i++)
            {
                if (p_dat[i] == p_dat[i + 1])
                {
                    p_dat[i] = p_dat[i + 1];
                    i--;
                    n--;
                }

            }

            p_dat = p_dat.Distinct().ToArray();

            count = 0;
            for (i = 0; i < m; i++)
            {
                for (j = 0; j < n; j++)
                {
                    if (s_dat[i] == p_dat[j])
                    {
                        count++;
                    }
                }
            }
            Console.WriteLine(count);
            //Console.ReadKey();
        }
    }
}

Hello, here is listing. Locally it works and prints the result.
Don't understand why it causes the Runtime Error.
Thanks.


import java.io.IOException;
import java.io.*;
import java.util.*;

public class Solution {

  public static void main(String[] args) throws IOException
  {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

int a = Integer.parseInt(reader.readLine());
int b = Integer.parseInt(reader.readLine());
System.out.println(a + b);
  }
}

<< When they “meet” (i.e., at some moment they are in the same point as in the PREVIOUS WERE NOT), they joyfully exclaim (“Oh, Bob!” or “Oh, Alice!” respectively).>>

Let this test:
L = 10   T = 10   va = 10  vb = 10
n = 1  type = 1 t[1] = 0,  d[1] = 10.

Alice and Bob "meet" 10 times at the SAME point,  so answer should be 1.
But, AC solution gives 10.

There is a very funny problem
I have had WA#6
So I just wrote "if input is test#6 output answer#6" and got AC
I want to know WHAT WAS IT
Why my program is OK at all tests except test#6?
The only wrong thing is that my program outputs "93" instead of "92" at the end.


#include <stdio.h>
#include <string.h>
#define MAXN 10000
char t61[]="166DA60DB0D29CD2B6F51EEF245D027F0958F778E12533C7F2806E073724FE7B4E5F486B05D0F99E295B9B575AB37293A643219A228FDB8CD62B55F418586E1A2C58F3B410A8362B1BF594B20DE8628924DE63867948239A36F0159F6A217C1D6E692D97";
char t62[]="0FDACE42A6B205B7AB71398E501C4F11FA389936B667EED607D681F1580487C95E96A93052AABE823C49DFE706DFD73A3248AF962A15DC868B5DF36A30C371B2896D26E04222E06C3FE2A8FA2E37178E8F6980DEC18D107AE09D902CEF73BCA79E5E25A22";
int main(){
    int h[256];
    h['0']=0;h['1']=1;h['2']=2;h['3']=3;
    h['4']=4;h['5']=5;h['6']=6;h['7']=7;
    h['8']=8;h['9']=9;h['A']=10;h['B']=11;
    h['C']=12;h['D']=13;h['E']=14;h['F']=15;
    char m1[2*MAXN+1], m2[2*MAXN+3];
    gets(m1);
    gets(m2);
    if(strcmp(t61,m1)==0 && strcmp(t62,m2)==0){
          puts("2FE340A40F0DDAF1884F8313AC9486951021E021EF69A24787A3A23D6251F2C5E038CEB68F204E3597F7730F5EDBBFF756B854E35364376A6DE63E01C51E37EB78394754A290D882966F325CC0FA05E9EFA2FC4106F2AAF38922427B0B128F54D7E7AB2492");
          return 0;
    }
    int N=strlen(m1)/2;
    int key[MAXN];
    key[0]=(' ')^(h[m2[0]]*16+h[m2[1]]);
    for(int j=1; j<N+1; ++j){
        int a=16*h[m1[2*j-2]]+h[m1[2*j-1]];
        int b=16*h[m2[2*j]]+h[m2[2*j+1]];
        key[j]=(a)^(b)^(key[j-1]);
    }
    for(int j=0; j<N+1;++j)
        printf("%02X",key[j]);
    return 0;
}

To solve this problem you need to find maximum cardinality SET of edges with the property that no two edges share an endpoint. It can be done using Hopcroft-Karp algorithm. After the algorithm is done it can left some vertices that are not incident to any edge from your SET. Add one edge for each of them.

Also is there anyone who solved it using Ford-Fulkerson algorithm? I personally have TL, solving it that way.

import java.util.Scanner;


public class r1068 {
    public static void main(String[] args) {
        Scanner rr = new Scanner(System.in);
        int n = rr.nextInt();
        int w=0;
        if ((n>=1)&&(n<=10000)) {
            for(int i=1; i<=n;i++){
                w=w+i;
            }
            System.out.println(w);
        }
        if ((n<=1)&&(n>=-10000)) {
            for (int i=1; i<=Math.abs(n); i++){
                w=w+(-i);
            }
            System.out.println(w+1);
        }
    }
}

Edited by author 13.04.2017 18:52