http://mendo.mk/jforum/recentTopics/list.page JForum - http://www.jforum.net http://cs.org.mk/index.php/19-nasloven-nastan/490-ciklus-natprevari-2022-start

Повеќе информации за натпреварите за основно и средно образование, може да се најдат на веб-сајтот на ЗИМ или на почетната страница на МЕНДО.]]> http://mendo.mk/jforum/posts/preList/718/3850.page http://mendo.mk/jforum/posts/preList/718/3850.page GMT
Повеќе информации за натпреварите за основно и средно образование, може да се најдат на веб-сајтот на ЗИМ или на почетната страница на МЕНДО.]]> http://mendo.mk/jforum/posts/preList/717/3849.page http://mendo.mk/jforum/posts/preList/717/3849.page GMT Internet online contest series

Over the next six months we will organize seven online contests as a warm-up for the 2021 season of high school programming competitions. Everyone is welcome to participate! Each contest will be three hours long and will feature five tasks. The contestants will also have full feedback with at most 50 submissions per task. The tasks will be of widely varying difficulty; we hope that many beginner or up-and-coming contestants will participate, as well as the more experienced ones.

The first contest will be held this Saturday, October 17, starting at 14:00 (GMT/UTC). Check out your local times at https://hsin.hr/coci/next_contest.html
You may use Python, Pascal, C/C++ or Java as your programming language of choice.
The contest will also be featured on a newly-developed judging system. If your students wish to get a bit more familiar with the system, we have uploaded the last 14 COCI seasons for them to practice on. Most problems (except for the very easy ones) have English translations.


The two relevant websites are:
https://hsin.hr/coci/ - information about the contest
https://evaluator.hsin.hr/ - judging system

We hope that you will join us or encourage your students to do so!
This is the fifteenth year in a row that we are hosting the COCI series. You can find the tasks (statements, test data and solutions) from the previous years at https://hsin.hr/coci/. There are over 600 original tasks for students to practice on!

With regards,
Kresimir Malnar
Croatian Computer Science Association]]> http://mendo.mk/jforum/posts/preList/716/3848.page http://mendo.mk/jforum/posts/preList/716/3848.page GMT
Повеќе информации можете да најдете на веб-сајтот на ЗИМ:
http://cs.org.mk/index.php/natprevari/srednoobrazovanie/makedonskaolimpijada-informatika

Задачите и тест случаите се веќе објавени во делот за тренинг на МЕНДО.]]> http://mendo.mk/jforum/posts/preList/715/3847.page http://mendo.mk/jforum/posts/preList/715/3847.page GMT http://mendo.mk/jforum/posts/preList/713/3843.page http://mendo.mk/jforum/posts/preList/713/3843.page GMT
USACO will organize several online contests in 2019 and 2020. Participation is free, and open to all. The contests are available in four divisions: bronze, silver, gold, and platinum.
All returning participants start out in their former divisions, and new participants start out in the bronze division. The contests are for individuals only, not teams.

Each contest has 3 problems to which you will submit solution programs in C, C++, Pascal, Java, or Python.
The USACO 2019 December contest is available from December 13 through December 16. Visit the USACO website for more information, including the schedule for future competitions: [url]http://usaco.org/[/url]
]]> http://mendo.mk/jforum/posts/preList/712/3842.page http://mendo.mk/jforum/posts/preList/712/3842.page GMT Internet online contest series
Direct link: [url]http://hsin.hr/coci/[/url]

Over the next six months we will organize seven online contests as a warm-up for the 2020 season of high school programming competitions. Everyone is welcome to participate!
Each contest will be three hours long and will feature five tasks. The tasks will be of widely varying difficulty; we hope that many beginner or up-and-coming contestants will participate, as well as the more experienced ones.

The first contest will be held on Saturday, 19 October 2019, starting at 14:00 (GMT/UTC). You may use Python, Pascal, C/C++ or Java as your programming language of choice.

Instructions, problems and the contest system will be located at [url]http://evaluator.hsin.hr[/url]

This web page will be available for registration 48 hours before the contest.
You can find problems, test data and solutions from previous Croatian competitions in informatics here: [url]http://hsin.hr/coci/[/url]
]]> http://mendo.mk/jforum/posts/preList/711/3841.page http://mendo.mk/jforum/posts/preList/711/3841.page GMT
Kako korisnici na sistemot, dokolku zabelezhite bilo kakov problem so MENDO, mozhete da ni ispratite e-mail poraka koristejki gi slednite kontakti: [url]http://mendo.mk/Help.do[/url]]]> http://mendo.mk/jforum/posts/preList/710/3840.page http://mendo.mk/jforum/posts/preList/710/3840.page GMT
JOI Open Contest 2019 is an IOI-like open competition for students at
schools for secondary education. The main purpose of this contest is
to give Japanese delegations and candidates of delegations an
opportunity for training and practice. But the contest itself is open to everybody.
Everybody is welcome to attend JOI Open Contest 2019!

This year, the contest will be held on 14 July 2019.
There will be 3 tasks. Each submitted source program must be written
in C++ (C++14). Problem statements will be provided both in Japanese
and English.

The duration of the contest is 5 hours. The same contests will be held
twice (Round 1 and Round 2). Contestants can participate in one of
Round 1 or Round 2 according to their time zone. After Round 2 finishes,
we will open the judging server for one day. Interested contestants can
improve and resubmit their solutions by themselves.

After the contest finishes, we will fix the standings.
No prizes will be given to the contestants.

Contest website:
[url]https://contests.ioi-jp.org/open-2019/index.html[/url]

Contest duration: 5 hours
The number of tasks: 3 tasks
Language: English, Japanese
Programming Language: C++ (C++14)
Details will be announced on the contest website.


Date & Time:
Round 1
Sunday, July 14, 2019
13:00-18:00 +0900 (JST)
04:00-09:00 (UTC/GMT)

Round 2 (tasks for Round 1 and Round 2 are the same)
Sunday, July 14, 2019
19:00-24:00 +0900 (JST)
10:00-15:00 (UTC/GMT)

Judging Server is open until
Monday, July 15, 2019
23:00 +0900 (JST)
14:00 (UTC/GMT)

======
Executive director of the Japanese Committee for the IOI
Seiichi Tani

Chairperson of the Scientific Committee of JCIOI]]> http://mendo.mk/jforum/posts/preList/709/3839.page http://mendo.mk/jforum/posts/preList/709/3839.page GMT [code]#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int,int> pi;
int n,m;
char mat[51][51];
bool vis[51][51];
vector<pair><int,int> > v1,v2,v3,e;
int di[4]={0,0,-1,1};
int dj[4]={1,-1,0,0};

bool valid_move(int i, int j)
{
if(i<0 || i>=n || j<0 || j>=m || mat[i][j]=='#')return false; // dali e ok potegot
return true;
}

bool to_the_end(vector<pi> v)
{

for(int i=0; i<v.size(); i++) >
{
if(v[i].first==e[i].first && v[i].second==e[i].second)continue; // dali stignav do F-s
else return false;
}
return true;
}
int bfs(vector<pi> start,vector<pi> end)
{
for(int i=0; i<n; i++) >
{
for(int j=0; j<m; j++) >
{
vis[i][j]=false;
}
}
queue<vector><pair><int,int> > > q;
vector<pair><int,int> > v5;
for(pi curr :start)
{
int i=curr.first;
int j=curr.second; //mu pravi true na site od pocetniot klavir
vis[i][j]=true;
v5.push_back({i,j});
}
q.push(v5);
queue<int> cekor;
cekor.push(0);
while(!q.empty())
{
vector<pi> v=q.front();
q.pop();
int c=cekor.front();
cekor.pop();
if(to_the_end(v))return c; // ako stignav do kraj
for(int k=0; k<4; k++)
{
vector<pi> nov;
bool flag=false;
for(int z=0; z<v.size(); z++) >
{
nov.push_back({v[z].first+di[k],v[z].second+dj[k]}); // go pomestuva cel objekt na nekoja nasoka
if(vis[v[z].first+di[k]][v[z].second+dj[k]]==true)flag=true;
}
bool can=true;
for(int z=0; z<nov.size(); z++) >
{
if(!valid_move(nov[z].first,nov[z].second)) // proveruva dali validna nasokata
{
can=false;
break;
}
}
if(can && !flag)
{
q.push(nov);
cekor.push(c+1);
for(int i=0; i<nov.size(); i++) >
{
vis[nov[i].first][nov[i].second]=true; // ako moze go zemam vektorot
}
}
}
}
return -1;

}
int main()
{
cin>>n>>m;
for(int i=0; i<n; i++) >
{
for(int j=0; j<m; j++) >
{

cin>>mat[i][j];
if(mat[i][j]=='1')
{
v1.push_back({i,j});
}
else if(mat[i][j]=='2')
{
v2.push_back({i,j});
}
else if(mat[i][j]=='3')
{
v3.push_back({i,j});
}
else if(mat[i][j]=='F')
{
e.push_back({i,j});
}
}
}
vector<int> res;
res.push_back(bfs(v1,e));
if(v2.size()>0)res.push_back(bfs(v2,e));
if(v3.size()>0)res.push_back(bfs(v3,e));
int ret=1000000;
int type=1;
for(int i=0; i<res.size(); i++) >
{
if(res[i]==0 || res[i]==-1)continue;
if(res[i]<ret) >
{
ret=res[i];
type=i+1;
}
}
if(ret==1000000)cout<<-1<<endl; >
else cout<<type><<endl<<ret><<endl;
return 0;
}
[/code]]]> http://mendo.mk/jforum/posts/preList/708/3836.page http://mendo.mk/jforum/posts/preList/708/3836.page GMT #include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int>arr;
void leftRotatebyOne()
{
int temp = arr[0], i;
for (i = 0; i < arr.size() - 1; i++)
arr[i] = arr[i + 1];

arr[i] = temp;

return;
}
void count_positive_rotations()
{
int good_rotations=0;
for(int j=0;j<arr.size();j++) >
{
int sum=arr[0];
bool good_rotation=true;
if(sum<0)
good_rotation=false;
else
{
for(int i=1;i<arr.size();i++) >
{
if(sum+arr[i]>=0)
sum+=arr[i];
else{
good_rotation=false;
break;
}
}
}
if(good_rotation)
good_rotations++;

leftRotatebyOne();
}


cout<<good_rotations><<endl;
return;
}
int main()
{
int N;
cin>>N;
for(int i=0;i<N;i++) >
{
int a;
cin>>a;
arr.push_back(a);
}
count_positive_rotations();

return 0;
}[/code]
????? ?? ?????? ?? ?????? 8 ????????, ?? ?????????? ?? ???? ?? ?????, ???? ?? ?? ???????]]> http://mendo.mk/jforum/posts/preList/707/3834.page http://mendo.mk/jforum/posts/preList/707/3834.page GMT
Немам никаква друга идеја за решавање на оваа задача освен со рекурзија(backtracking)
каде пробувам секој број од едната низа да го сведам во број од другата низа, а потоа рекурзивно решавам за следната состојба

Поконкретно, паѓа на овој тест пример

9
1 14 22 11 6 11 14 1 24
5
34 37 20 8 5

каде:

очекувано: 8
добиено: INT_MAX(не наоѓа решение)

Еве го кодот
[code]
#include <iostream>
#include <bits/stdc++.h>

using namespace std;

// Sorts the vectors and removes elements that are contained in
// both s and f
// if we try to solve for s = {2, 3, 7} f = {2, 4, 6}
// it is same as solving s = {3, 7} f = {4, 6}
void prepare(vector<int> &s, vector<int> &f)
{
sort(s.begin(), s.end());
sort(f.begin(), f.end());

vector<int> toDelete;
for(int i = 0; i < s.size(); i++)
{
for(int j = 0; j < f.size(); j++)
{
if(f[j] == s[i])
{
f.erase(f.begin() + j);
toDelete.push_back(s[i]);
break;
}
}
}

for(int i = 0; i < toDelete.size(); i++)
{
s.erase(find(s.begin(), s.end(), toDelete[i]));
}
}

int solve(vector<int> &s, vector<int> &f)
{
prepare(s, f);

// special cases that can be treated as base cases
// the first case: if we have to convert s = {18} to f = {2,3,6,7} we need to decompose 18
// to get s = {2, 16} f = {2, 3, 6 ,7}, then decompose 16 to get s = {2, 3, 13} f = {2, 3, 6, 7}
// similar situation for the second case
if(s.size() == 1)
return f.size() - 1;
if(f.size() == 1)
return s.size() - 1;

int result = INT_MAX - 10;

for(int i = 0; i < s.size(); i++)
{
for(int j = 0; j < f.size(); j++)
{
// try to decompose ith element of s if s[i] > f[j]
if(s[i] > f[j])
{
vector<int> ss = s;
vector<int> ff = f;

ss.erase(ss.begin() + i);

ss.push_back(f[j]);
ss.push_back(s[i] - f[j]);

result = min(result, solve(ss, ff) + 1);
}
// try to find another number x where s[i] + x equals to f[j]
else if(s[i] < f[j])
{
auto x = find(s.begin() + i+1, s.end(), f[j] - s[i]);
if(x != s.end())
{
vector<int> ss = s;
vector<int> ff = f;

ss[i] = f[j];
auto doomed = find(ss.begin() + i+1, ss.end(), f[j] - s[i]);
ss.erase(doomed);

result = min(result, solve(ss, ff) + 1);
}
}
}
}

return result;
}

int main()
{
int N;
cin >> N;
vector<int> s;
int sumS = 0;
for(int i = 0; i < N; i++)
{
int si;
cin >> si;
sumS += 0;
s.push_back(si);
}

int M;
cin >> M;
vector<int> f;
int sumF = 0;
for(int i = 0; i < M; i++)
{
int fi;
cin >> fi;
sumF += 0;
f.push_back(fi);
}

if(sumS != sumF)
{
cout << -1;
}
else
{
cout << solve(s, f);
}

return 0;
}

[/code]

Забелешка: На МЕНДО 9/20 се точни а останатите се погрешен резултат или надминат временски лимит]]> http://mendo.mk/jforum/posts/preList/705/3827.page http://mendo.mk/jforum/posts/preList/705/3827.page GMT #include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
long long N, K;
cin>>N>>K;
vector<long long>num;
for(long long i=0;i<N;i++) >
{
long long a;
cin>>a;
num.push_back(a);
}
long long rez=0;
sort(num.begin(), num.end());
for(long long i=0;i<N;i++) >
{
for(long long y=i+1;y<N;y++) >
{
if(num[i]+K==num[y])
rez++;
}
}
cout<<rez><<endl;
return 0;
}
[/code]
Со кодов ми поминува на 27/37 тест примери на останатите ми паѓа на време, дали со неколку модификации на кодов може да се реши задачава или ќе треба да го земам фактот дека 1≤N≤100 000, а дека задачава работи О(N na kvadrat)]]> http://mendo.mk/jforum/posts/preList/704/3818.page http://mendo.mk/jforum/posts/preList/704/3818.page GMT Fala odnapred :D
]]> http://mendo.mk/jforum/posts/preList/703/3817.page http://mendo.mk/jforum/posts/preList/703/3817.page GMT #include <iostream>
#include <string>
using namespace std;

int main()
{
float c;
string p;
cin >> p >> c;
int br, pro=0;
cin >> br;
if(c>=1 && c<=1000 && br>=1 && br<=50) {
string niza[br];
for(int j=1; j<=br ;j++) {
cin >> niza[j];
}
for(int i=1; i<=br ;i++) {
if(p==niza[i]) {
pro=1;
}
}
if(pro==1) {
c=c+c/20;
} else {
c=c+c*18/100;
}
cout << c;
}
return 0;
}
mi prikazuva runtime error]]> http://mendo.mk/jforum/posts/preList/702/3815.page http://mendo.mk/jforum/posts/preList/702/3815.page GMT