#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

int main(){
    int k;
    cin>>k;
    vector<int> line1;
    vector<int> line2;
    vector<int> sum;
    while(k--){
        int m,n;
        cin>>m>>n;
        line1.push_back(m);
        line2.push_back(n);
    }
    bool plus_1=false;
    for(int i=line1.size()-1;i>=0;i--){
        int temp=line1[i]+line2[i];
        if(plus_1){
            temp++;
        }
        if(temp>=10){
            plus_1=true;
            temp-=10;
        }else{
            plus_1=false;
        }
        sum.push_back(temp);

    }
    for(vector<int>::reverse_iterator iter=sum.rbegin();iter!=sum.rend();++iter){
        cout<<*iter;
    }
return 0;
}

I've been wa on this case for a long time:
abaabaaba
The right answer is abaabaabaaba

Can anybody please give me cases for Test2 ?

sandro
answer is 5

Really fast.

for (int k = 0; k < n; ++k) {
  for (int i = graph[k]._Find_first(); i < n; i = graph[k]._Find_next(i)) {
    graph[k] |= graph[i];
  }
}

I created array p[1..10000000] of boolean; p[i]=true when exist i in input. But i can t prove that if length of input <=100000 then absent numb <=10000000

Edited by author 11.02.2019 20:58

How can I get this?

All compilers are succeed with this code.
What's wrong?


#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()

{
  double buffer[128*1024];
  unsigned long long number, n=0;
  while(cin >> number)
  {
    buffer[n]=pow(number,1.0/2.0);
    n++;
  }
  for (int i=n-1;i>=0;i--)
  {
    cout << fixed << setprecision(4) << buffer[i] << endl;
  }
  return 0;
}

So for this test:

0 0 90 5
0 -0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

correct answer is 31.4159, right?


Edited by author 01.02.2019 23:40

It's very funny to read how I tried to solve:

1-st time I forgot to multiply answer by 2 :)
2-nd time I retyped code by forgot to set 'Pascal' (and it was compile error)
3-rd time I forgot 'begin' :)
4-th time I typed task correctly and it was accepted !!!

So, programmers, DO NOT HURRY if you don't want to repeat my funny story :)

Edited by author 10.09.2004 00:28

the difficulty rangering here is absolutely uncorrect and unjust

Edited by author 31.01.2019 04:21

I used template programming to write 3D Fenwick tree and I get "Compiler failed" verdict when compiling with G++ (not "Compilation Error", but "Compiler failed"; no error message). The solution works fine and gets Accepted with Visual Studio and Clang.

I assume it is because of the recursive template operations, but it works when I compile it locally with G++ (using the flags provided on help.aspx?topic=cpp). It also work on other online judges, such as codeforces, which also use mingw g++.

Could I get some insight on this? I'm curious why this happens.

#include <iostream>
using namespace std;
int main() {
    long x, y, c, a, b, MIN, MAX;
    cin >> x >> y >> c;
    if (x + y < c) {
        cout << "Impossible";
        return 0;
    }
    if (x >= c)
        cout << c << " " << 0;
    else if (y >= c)
        cout << 0 << " " << c;
    else{
        if (x < y)
            cout << x << " " << c - x;
        else
            cout << c - y <<  " " << y ;
    }
}

This is my code.
#include <iostream>
#include <vector>
using namespace std;
int main()
{
    int n,a,k=0;
    cin>>n;
    vector <int> v;
    vector <int> sum;
    vector <int> index;
    for(int i=0; i<n; i++)
    {
        cin>>a;
        v.push_back(a);
    }
    for(int i=1; i<n-1; i++)
    {
        sum.push_back(v[i-1]+v[i]+v[i+1]);
        index.push_back(i);
    }
    int max=sum[0];
    for(int i=1; i<n; i++)
    {
        if(sum[i]>max) {max=sum[i]; k=i;}
    }
    cout<<max<<" "<<index[k]+1;
    system("pause");
    return 0;
}

just be patient while making precalculation :)

Жеванный крот, одна халтура! Мало того что английский перевод кривой как рог барана, так ещё и проверяющая система хромает!

Отправляем на проверку:
Var
s, s1: ansistring;
a, b: byte;

Begin
readln(s, s1); val(s[length(s)], b);

End.

Получаем вполне обоснованный WA1.
Теперь дополним:
Var
s, s1: ansistring;
a, b: byte;

Begin
readln(s, s1); val(s[length(s)], b);
val(s1[length(s1)], a);

End.

Всё, вселенная дрогнула: "Access violation"... *facepalm*
И всё бы ничего, но подобное происходит уже где-то раз пятый (в разных задачах).
И да, прочитать string целиком, а затем разрезать - выдаёт ту же ошибку AV:
Var
s, s1: ansistring;
i, a, b: byte;

Begin
readln(s); for i:=1 to length(s) do if s[i]:=' ' then begin val(s[i-1], b); val(s[length(s)], a); break; end;

End.

Сделайте уже нормальный ввод!!!!1

Edited by author 01.02.2020 21:32

I recommend you this resource: https://e-maxx.ru/algo/bfs
also you can make a function instead of the lyambda, but I just like lyambdas more.
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
int main() {
    int N;
    cin >> N;
    vector <vector <int>> g(N);
    int h;
    for (int i = 0; i < N; i++) {
        for (;;) {
            cin >> h;
            if (!h)
                break;
            g[i].push_back(h - 1);
        }
    }
    int n = g.size();
    queue<int> q;
    vector<bool> used(n);
    vector<int> d(n);
    int s;
    auto bfs = [&](int start) {
        s = start;
        q.push(s);
        used[s] = true;
        while (!q.empty()) {
            int v = q.front();
            q.pop();
            for (size_t i = 0; i<g[v].size(); ++i) {
                int to = g[v][i];
                if (!used[to]) {
                    used[to] = true;
                    q.push(to);
                    d[to] = d[v] + 1;
                }
            }
        }
    };
    for (int start = 0; start < n; ++start)
        if (!used[start]) {
            bfs(start);
        }
    vector <int> ans;
    for (int i = 0; i < n; i++) {
        if (d[i] % 2 == 0)
            ans.push_back(i + 1);
    }
    int size = ans.size();
    cout << size << '\n';
    for (int i = 0; i < size; i++)
        cout << ans[i] << " ";
}

Edited by author 26.01.2019 22:53

Edited by author 30.01.2019 03:16