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

string S, A;
unsigned short An = 1;
bool negative     = true;

void get_next_A();
void get_S(unsigned short);

int main()
{
    unsigned short input_n;

    cin >> input_n;

    get_S(input_n);

    cout << S << endl;

    return 0;
}

void get_next_A()
{
    if(An == 1)
        A = "sin(1)";
    else
    {
        A.erase(A.size() - An + 1, An);

        if(negative)
            A += "-sin(";
        else
            A += "+sin(";

        char buffer[5];

        itoa(An, buffer, 10);
        A += buffer;

        for(unsigned short j = 0; j < An; ++j)
            A += ")";

        negative = !negative;

    }

    ++An;
    return;
}

void get_S(unsigned short n)
{
    char _buffer[10];

    for(unsigned short i = 0; i < n - 1; ++i)
        S += '(';

    for(unsigned short i = n; i > 0; --i)
    {
        get_next_A();

        itoa(i, _buffer, 10);

        S += A;
        S += '+';
        S += _buffer;
        if(i != 1)
            S += ')';

    }

    return;
}

#include <iostream>
using namespace std;
int main() {
    int n,res=0;
    cin >> n;
    if (n < 0) {
        n = n*-1;
        int *arr;
        arr = new int[n+2];
        arr[0] = 1;
        for (int i = 1; i < n+2; i++) {
            arr[i] = arr[i - 1] - 1;
        }
        for (int i = 0; i < n+2; i++) {
            res = res + arr[i];
        }
        cout << res;
    }
    else if(){
        int *arr;
        arr = new int[n];
        arr[0] = 1;
        for (int i = 1; i < n; i++) {
            arr[i] = arr[i - 1] + 1;
        }
        for (int i = 0; i < n; i++) {
            res = res + arr[i];
        }
        cout << res;
    }
    return 0;
}

program Suma;
var a,b,e:integer;
begin
read(a);
b:=1;
if a<0 then
e:=((1+a) div 2)*(((a-1)div((b-1)-1))+1)
else
e:=(a*(a+1)) div 2;
writeln(e);
end.

var i,n,q,h:integer;
k:real;
begin
q:=0;
read(n);
h:= 10000;
if (n <= h) and (n >= -h) then begin
if n>0 then begin
k:= ((1+n)/2)*n;
writeln(k);
end;
if n=0 then begin
k:=0+1;
writeln(k);
end;

if n<0 then begin
for i:= 1 downto n do
inc(q);
k:= ((1+n)/2)*q;
writeln (k);
end;
end;
end.

I tried binary search, harsh table, got TLE. then I thought using direct mapping table could get MLE, but never TLE, so i tried it, still TLE!!!tell me how could this code get TLE.
#include <iostream>
using namespace std;
char a[1000000001];
int main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        int j;
        cin >> j;
        a[j] = 1;
    }
    int m;
    cin >> m;
    int cnt= 0;
    for (int i = 1; i <= m; ++i) {
        int j;
        cin >> j;
        cnt += a[j];
    }
    cout << cnt;
}

#include <stdio.h>
int main(){
    int a, b,c;
    scanf_s("%d",&a);
    scanf_s("%d",&b);
    c = a + b;
    return 0;
}

//the result is wrong answer,I want to ask why?Can you help me fix it?

check whether months have higher priority than days in your program


Edited by author 20.10.2016 20:42

big simulation spend one day+ and 700+ lines without seeing test data

It's not a programming challenge.
It's more of "Can you read me whole?" challenge.

Kinda reminds License agreement...

В примере, в результате написано "1 1", однако в вычислениях мы видим "1 2 - 2 1 - 0 3". Откуда, спрашивается, взялась еще одна единица? Каждый собрал по 1 кг, при пересыпаниях туда-сюда получаем 0кг и 2кг и наоборот. Объясните нормально условие

If K=8,
i.e 1,2,3,4,5,6,7,8 students.
Now, each student has to sit for at least 1 lab of each professor.

For N=11,
Students 1 will sit in all labs, hence qualified for exam.
Students 2 will sit in labs: 2,4,6,8,10,12,14,16,18,20,22. hence qualified for exam. (At least 1 lab of all Prof covered)
Students 3 will sit in labs: 3,6,9,12,15,18,21,24,27,30,33 labs. hence qualified for exam. (At least 1 lab of all Prof covered)
.
.
.
Similarly Student 8 will sit in 8,16,.....,88 labs, hence qualified for exam.  (At least 1 lab of all Prof covered)



Please help, where I'm wrong.

If N = 0 Q = -1 or Q = 0?

I use LINQ to sort this sequence, it works pretty well on all tests i found.
Any ideas why it works wrong on test 4?



using System;
using System.Linq;

namespace Таблица_результатов
{
    class Program
    {

        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());
            string[] lines = new string[n];
            for (int i = 0; i < n; i++) { lines[i] = Console.ReadLine(); }

            var a = lines
                 .Select(x => x.Split())
                 .ToArray()
                 .OrderByDescending(x => x[1]);


            foreach (var x in a)
            {
                Console.WriteLine(x[0] + " " + x[1]);
            }
        }
    }
}

another easy 10000+ problem..haha

don't use hashes with modulo = 2^32, change it to 10^9 + 7

Hi, i have problem with WA#52, I used long long and test 1 1 32 => accept.
This's my code :
#include <iostream>
using namespace System;
using namespace std;
long long func( long long x , int po );

void main() {

    long long a,b,c,y,k,x,n,m;
    cin>>a>>b>>c;

    k = (c-1)/a;
    m = (c-1)/b;
    x = func( k ,2 );
    y = func( m , 2 );

    printf("%lld\n%lld\n2",x,y);
   system("pause");
}

long long func( long long x , int po ) {
    long long temp=2;
    for( int i=1; i<x; i++ ) {
        temp *= po;
    }
    return temp;
}

It's not so interesting to publish ready code, but for those who stuck on this problem the following algorithm could be useful:

1. First of all, build generator, that will generate random tests with specified phone number length and number of words. It will helpful, since you don't know, which tests are runned when you send you code for check.

2. Convert your words to digits and remove those of them that aren't contained in phone number. Also remove duplications that can appear after this converting. During this process build dictionary that will contain converted word as a key, and original word as a value. Last one will be needed to output the result.

3. Sort converted words by length in descendent order.

4. Take every converted word and find all its inclusions to the phone number. Build dictionary that will contain found index as key and list of converted words as value.

5. Now use recursive function which initially will get index=0 as argument, find all words from dictionary for the index and loop them, taking their length to get next index.
If next index equals phone number length, you found one of the solutions.

6. Optimize the function to eleminate redundant calls. I mean, on specific step you can check if the next searched solution is expected to be better than found one or not.

I'm not sure that this algorithm is the best, but at least I got "accepted" on C# with 600 ms performance.

My approach divides into 2 cases

I) A>= sqrt(N) or B>=sqrt(N)
  this can be solved by brute force in O(sqrt(N))

II) A<sqrt(N) and B<sqrt(N)
  now this is more interesting
  in case of gcd(A,B)=1
  you can proof that there exist x,y>=0 that A*x+B*y = N
    since A*x congruent to N (mod B) -> x = N*inv(A,B) % B
    hence 0<=x<=B-1 but B<sqrt(N) and A<sqrt(N)
    then A*x < N so y >= 0
  in case of gcd(A,B)>1
    that's your work!

:)

My program answer true for input from example(for big number too), but i get "Wrong answer" on test 1.

#include <stdio.h>
#include <math.h>
#define N 1000000
int main(int argc, char **argv)
{
    long long int array[N];
    int i=0;
    while (scanf("%Ld", &array[i])>0) {
        i++;
    }
    for(i-=1;i>=0;i--)
        printf("%.4f\n", sqrt(array[i]));
    return 0;
}

I don't think problem statement is clear.. and obviously algorithm difficult is far not worth than 10000+ reading comprehension is possible

Edited by author 16.10.2016 11:24

Don't use scanf !!!!!!

Edited by author 16.10.2016 18:23