If using recursive algorithm increase recursion depth limit at least 4000. For my algo 2000 was not enough

import sys
sys.setrecursionlimit(4000)

>>> print(round(0.0))
>>> 0.0
But
"The output should contain the required distance in meters rounded to two fractional digits"
So don't forget to use string formatting to provide 2 digits after comma

Real test case:
5 90 2.50
0.00

This is proof of why answer < 3.

Suppose, we have 3 elements x, y, z, with gcd(x, y, z) != 1. Then, after first transformation array will contain elements: gcd(x, y), gcd(x, z), gcd(y, z), Let's call them a = gcd(x, y), b = gcd(x, z), c = gcd(y, z). After second tranformation array will contain elements: gcd(a, b), gcd(a, c), gcd(b, c), we can notice, that:
gcd(a, b) = gcd(gcd(x, y), gcd(x, z)) = gcd(x, y, x, z) = gcd(x, y, z). Doing similar things with gcd(a, c) and gcd(b, c), it turns out, that gcd(a, b) = gcd(a, c) = gcd(b, c) = gcd(x, y, z). That means, that after 2nd transformation array will contain 3 equal numbers greater than 1. I think it's obvious, that answer will be infinity, if array contains 3 equal numbers greater than 1.


Now, if gcd of any tripple of numbers equals one, that means, that we will end in less than 2 transformations (Proof this by yourself)

Edited by author 22.01.2022 15:49

var
  a: array[1..2000] of string;
  i: integer;
begin

  for i := 1 to 2000 do
  begin
    if (i >= 1) and (i < 5) then a[i] := 'few';
    if (i >= 5) and (i < 10) then a[i] := 'several';
    if (i >= 10) and (i < 20) then a[i] := 'pack';
    if (i >= 20) and (i < 50) then a[i] := 'lots';
    if (i >= 50) and (i < 100) then a[i] := 'horde';
    if (i >= 100) and (i < 250) then a[i] := 'throng';
    if (i >= 250) and (i < 500) then a[i] := 'swarm';
    if (i >= 500) and (i < 1000) then a[i] := 'zounds';
    if i >= 1000 then a[i] := 'legion';
  end;

  readln(i);
  writeln(a[i]);

end.

Edited by author 05.12.2007 18:27

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;
}