Your memory limits are too tight for 64-bit compilers and using 32-bit Visual C++ instead is not such a good idea because of its weird behavior.

Rust releases new version every 6 weeks. Even if this is too frequent please update it at least every half year

Test #10 contains some non utf-8 characters.
That could be the reason of panic if you use Go or Rust languages.

Just pay attention when Conviviality rating < 0.

floor(x) or round(x) - wa7
(int)x - OK

Can anybody describe the solution please ?

I noticed a lot of people complain about WA2. Just wanted to share my experience, since I got WA2 several times. Don't forget that after you read the number, you haven't read the newline after it. So cin.getchar + cin.ignore and getline(cin, ...) get that exact newline.

So we must try to make teams have equal size. Why?
Consider two teams with x and y members respectively, and consider x > y. Sending a member from first team to second team will only affect matches between these two teams. At first we have x*y different matches between the teams. If we send a player to the second team, there are now x-1 and y+1 members in each team. The matches are now
(x-1)*(y+1) = x*y + (x-y) - 1
As we said x > y, x-y > 0, so x-y-1 >= 0 and our change can't reduce the total number of matches, so it is an optimal change.
This also shows that nothing changes when x = y+1, so if we can't make all teams equal size, just add each of the ones remaining into a different team, making a difference of at most 1 in the sizes of teams.

Now it's your job to compute the number of matches, I won't show that, but it can be done with a one line formula. Good luck :)

I have solved this problem by using an easy greedy algorithm.

MAIN: you need to use an idea of 2 pointers(of course you need to sort array) and take the rightest segment. Left part of segment need to be <= x(another pointer)

The solution below:

void solve() {
    int m;
    cin >> m;
    vector< pair<int, int> > pi;
    int a, b;
    cin >> a >> b;
    while (a || b) {
        pi.pb({a, b});
        cin >> a >> b;
    }
    sort(all(pi));
    int cnt = 0, j = 0;
    vector< pair<int, int> > ans;
    j = 0;
    int x = 0;
    for (; j < sz(pi) && x < m; j++) {
        int k = j;
        int mx = x;
        pair<int, int> par = {-INF, -INF};
        while (k < sz(pi) && pi[k].ff <= x) {
            if (mx < pi[k].ss) {
                mx = pi[k].ss;
                par = pi[k];
            }
            k++;
        }
        if (par.ff == -INF) {
            cout << "No solution\n";
            return;
        }
        x = mx;
        ans.pb(par);
        if (k > j)
            j = k - 1;
    }
    if (x < m) {
        cout << "No solution\n";
        return;
    }
    cout << sz(ans) << '\n';
    for (auto &i : ans) {
        cout << i.ff << ' ' << i.ss << '\n';
    }
}

I dont understand why difficulty is 121, i think it's easy problem

I had a hard time with this, so hopefully I will save your time. Read input like this:

int n; cin >> n;
string t;
getline(cin, t); // Necessary to process end of first line (nothing after n)
....
// Now keep getting each line with getline(cin, t)

Hints:

1. No sorting. It's useless. Quiqsort - too much memory, Bublesort - too long time.

2. We just need a number that is more than n/2. It always exists.

3. Read values step by step, remember the 'candidate' answer and counter increase or decrease it.

e.g:

3->3 cnt=0+1=1

3->3 cnt=1+1=2

2->3 cnt=2-1=1

2->3 cnt=1-1=0

3->3 cnt=0+1=1

Note 1:

e.g.

5

3

3

2

2

6

My program's answer is 6, but such a case never occurs (!).

"He knows exactly that more than half banknotes have the same value."

Note 2:

C# - "Memory limit exceeded 21"

use this:

    if (i % 1000 == 0)

       GC.Collect();

Regards

Edited by author 12.01.2022 20:54

using System;

namespace Timus
{
    class Program
    {
        static void Main()
        {
            string nums = Console.ReadLine();
            int a = Convert.ToInt32(nums.Substring(0, 1));
            int b = Convert.ToInt32(nums.Substring(2, 1));
            int c = a + b;
            Console.WriteLine(c);
        }
    }
}

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main ()
{    long long n;
    while (cin>>n)
{
    cout<<setprecision(4)<<fixed<<sqrt(n)<<endl;
}
return 0;
}

Edited by author 12.11.2017 17:07

This is my code!!!
#include<iostream>
#include<string>
using namespace std;
int main() {
 char s;
 int count = 0;
 while (cin >> s) {
    if ((s=='a')||
        (s=='d')||
        (s=='g')||
        (s=='j')||
        (s=='m')||
        (s=='p')||
        (s=='s')||
        (s=='v')||
        (s=='y')||
        (s=='.')||
        (s==' ')) {count+=1;}
    if ((s=='b')||
        (s=='e')||
        (s=='h')||
        (s=='k')||
        (s=='n')||
        (s=='q')||
        (s=='t')||
        (s=='w')||
        (s=='z')||
        (s==',')) {count+=2;}
    if ((s=='c')||
        (s=='f')||
        (s=='i')||
        (s=='l')||
        (s=='o')||
        (s=='r')||
        (s=='u')||
        (s=='x')||
        (s=='!')) {count+=3;}
 }
 cout << count;
system("pause");
 return 0;
}

Soln accepted but time is .156 how to reduce to atleast 0.078 and also how to reduce memory usuage?
import java.util.*;
import java.io.*;
public class prob1264{
public static void main(String args[])
{
    Scanner sc=new Scanner(System.in);
    PrintWriter out=new PrintWriter(System.out);
    int n=sc.nextInt();
    int m=sc.nextInt();
    int result=n*(m+1);
    out.print(result);
    out.close();
    }
}

program p_e;
const
  maxn=20000+1;
type
  tgroup=record father,members:integer;end;
  ttree=array[1..maxn]of tgroup;
  tstack=array[1..maxn]of integer;
  pnode=^tnode;
  tnode=record node:integer; next:pnode; end;
  plist=array[1..maxn]of pnode;
var
  n,k:longint;
  hands:longint;
  groups:ttree;
  ngroups:integer;
  stack:tstack;
  startstack,endstack:integer;
  list:plist;

procedure Add(x,y:integer);
var
  p:pnode;
begin
  new(p);
  p^.node:=y;
  p^.next:=list[x];
  list[x]:=p;
end;

procedure read_data;
var
  ng:integer;
  index:integer;
  i:integer;
  son,mem:integer;
begin
  readln(n,k);
  ngroups:=1;
  groups[1].father:=0;
  groups[1].members:=n;
  fillchar(list,sizeof(list),0);
  while not eof do
  begin
    read(index);
    read(ng);
    for i:=1 to ng do
    begin
      read(son,mem);
      Add(index,son);
      inc(ng);
      groups[son].father:=index;
      groups[son].members:=mem;
    end;
  end;
end;

procedure EliminateOneCouple(node:integer);
begin
  if node=0 then
    exit;
  dec(groups[node].members);
  EliminateOneCouple(groups[node].father);
end;

procedure eliminate_couples;
var
  p:pnode;
begin
  startstack:=1;
  endstack:=1;
  stack[1]:=1;
  while startstack<=endstack do
  begin
    p:=list[stack[startstack]];
    if (p=nil) and (p^.node=2) then
      EliminateOneCouple(stack[startstack])
    else
      while p<>nil do
      begin
        inc(endstack);
        stack[endstack]:=p^.node;
        p:=p^.next;
      end;
    inc(startstack);
  end;
end;

procedure get_handshackes;
var
  p:pnode;
  current_shackes:longint;
begin
  fillchar(stack,sizeof(stack),0);
  startstack:=1;
  endstack:=1;
  stack[1]:=1;
  while startstack<=endstack do
  begin
    p:=list[stack[startstack]];
    current_shackes:=1;
    if p=nil then current_shackes:=0;
    while p<>nil do
    begin
      inc(endstack);
      stack[endstack]:=p^.node;
      current_shackes:=current_shackes*groups[p^.node].members;
      p:=p^.next;
    end;
  inc(startstack);
  inc(hands,current_shackes);
  end;
end;

procedure write_hands;
begin
  writeln(hands);
end;

begin
  read_data;
  eliminate_couples;
  get_handshackes;
  write_hands;
end.

Try to reread the task. I've been adding up wrong numbers together because of the misunderstanding.

Also, try tests from "If you have WA6"

Why i get RuntimeError?

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;
using System.Threading;
using System.Diagnostics;
using System.Globalization;


namespace AlgorytmyInternetowe
{
    public class Program
    {

        public static void Main(string[] args)
        {
            //string[] tmp = Console.In.ReadToEnd().Split(new char[] { ' ' });
            string[] tmp = Console.ReadLine().Split(new char[] { ' ', '\t' }, StringSplitOptions.RemoveEmptyEntries);
            Thread.CurrentThread.CurrentCulture = CultureInfo.InvariantCulture;
            Console.Write(int.Parse(tmp[0])-int.Parse(tmp[4])+" "+ (int.Parse(tmp[1]) - int.Parse(tmp[3])));
        }
    }
}

A could do the simple solution there I have 4 points of square and simple calculations of dist to this square. But i found this very easy and i figure out difficult formula for equations of the lines which bounding this square. I calculates dist for this lines and analyzing them.

This is initialization of square:
/*
void init(Point p1, Point p2){
    int64_t dx = p1.x-p2.x, dy = p1.y-p2.y;
    a = dx + dy; // some coof for line-equation
    b = dx - dy;
    c1 = dx*(p1.x-p1.y) + dy*(p1.x+p1.y);
    c2 = dy*(p2.x-p2.y) - dx*(p2.x+p2.y);
    dc = dx*dx + dy*dy; // delta between c-coof two parallel lines
    if (dc == 0) {
        // if square is dot
        c1 = p1.x;
        c2 = p1.y;
   }
}
*/
There i calculate dists to lines
/*
if (dc == 0){
    return std::hypot((p0.x - c1), (p0.y - c2));
} else {
    double
    d1 = (b*(p0.y) - a*(p0.x) + c1)/sqrt(2*dc), // signed dist to first line
    d2 = (b*(p0.x) + a*(p0.y) + c2)/sqrt(2*dc), // signed dist to perpendicular to first line
    d = sqrt(dc/2.0); // delta between two parallel lines (a side of square)
......
*/
"d" always is positive (if square isn't dot), bcs "(d1 - d) < d1" (dist to nearest line less than dist to farthest line)



there is some conditionals for calculate the dist to square
/*
if (d1 < 0) {
    if (d2 < 0) {
        // 1 corner   x1 = (c1*a - c2*b)/(2.0*dc), y1 = (-c1*b - c2*a)/(2.0*dc)
    } else if(d2 <= d){
        // 1 side
        return fabs(d1);
    } else {
        // 2 corner   x2 = (c1*a - c2*b)/(2.0*dc) + (b/2.0), y2 = (-c1*b - c2*a)/(2.0*dc) + (a/2.0)
    }
} else if (d1 <= d) {
    if (d2 < 0){
        // 2 side
        return fabs(d2);
    } else if(d2 <= d){
        // inside square
        return 0;
    } else {
        // 3 side
        return fabs(d2 - d);
    }
} else {
    if (d2 < 0){
        // 3 corner   x3 = (c1*a - c2*b)/(2.0*dc) - (a/2.0), y3 = (-c1*b - c2*a)/(2.0*dc) + (b/2.0)
    } else if(d2 <= d){
        // 4 side
        return fabs(d1 - d);
    } else {
        // 4 corner   x4 = (c1*a-c2*b)/(2.0*dc) + (b-a)/2.0, y4 = (p0.y + (c1*b+c2*a)/(2.0*dc) + (a+b)/2.0
    }
}
*/
after this we sorting dist with num, and print the answer

tests:
8
0 0 2 2
2 4 0 6
-8 -3 -8 -7
7 11 4 15
5 -5 9 -6
-2 -2 -2 -4
-4 6 -8 10
6 6 10 6
.....
6 2 : ans is "8 1 2 5 6 4 7 3"
0 0 : ans is "1 6 2 5 7 3 8 4"

Edited by author 26.08.2020 02:41