Why do they want me to do?
How to understand "equal-sized parts"?

are leading zeroes allowed?
for example... will 0110 be counted as a valid 4 digit lucky number?

#include<bits/stdc++.h>
using namespace std;
//input section
#define si1(x)       scanf("%d",&x)               //one integer
#define si2(x,y)     scanf("%d%d",&x,&y)          //two integers
#define si3(x,y,z)   scanf("%d%d%d",&x,&y,&z)     //three integers
#define sl(x)        scanf("%lld",&x)             //long long int
#define sul(x)       scanf("%llu",&x)             //unsigned long long
#define ss(x)        scanf("%s",s)                //string
#define sd(x)        scanf("%lf",&x)              //double

//output section
#define   pf          printf
#define   pfi(x)      printf("%d\n",x)
#define   pfl(x)      printf("%lld\n",x)
#define   pfd(x)      printf("%lf\n",x)
#define   nl          printf("\n")

//data type
#define   ll          long long int
#define   ld          long double
#define   ull         unsigned long long

//general
#define    pb          push_back
#define    mp          make_pair
#define    f           first
#define    s           second
#define    all(x)      x.begin( ),x.end( )
#define    sv(x)       memset(x, 0, sizeof(x))
#define    PI          3.1415926535897932384626
#define    mod         1000000007

//loop
#define    lp(a,b)      for(int i=a;i<b;i++)            //assending order
#define    lpr(a,b)     for(int i=b;i>=a;i--)           //decending order
#define    pl(a)        pair<ll, ll>a
#define    ppi(a)       pair<int,int>a
#define    vec(v)       vector<int>v
#define    vi(v,itr)    vector<int>::iterator itr=v.begin( );

void solve( )
{
    int cnt=0;
    char s,h;
    cin>>s>>h;
    int p,q;
    p=(int)s-96;
    q=(int)h-48;
    if((p+1)>=1 && (p+1)<=8 && (q-2)>=1 && (q-2)<=8) ++cnt;
    if((p-1)>=1 && (p-1)<=8 && (q+2)>=1 && (q+2)<=8) ++cnt;
    if((p-1)>=1 && (p-1)<=8 && (q-2)>=1 && (q-2)<=8) ++cnt;
    if((p+1)>=1 && (p+1)<=8 && (q+2)>=1 && (q+2)<=8) ++cnt;
    if((p+2)>=1 && (p+2)<=8 && (q-1)>=1 && (q-1)<=8) ++cnt;
    if((p-2)>=1 && (p-2)<=8 && (q+1)>=1 && (q+1)<=8) ++cnt;
    if((p-2)>=1 && (p-2)<=8 && (q-1)>=1 && (q-1)<=8) ++cnt;
    if((p+2)>=1 && (p+2)<=8 && (q+1)>=1 && (q+1)<=8) ++cnt;
    cout<<cnt<<endl;
}

int main( )
{
    //code by noob coder BD (Shahariar,CSE,RUET-19)

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    int t=1;
    cin>>t;
    while(t--)
    {
        solve( );
    }
    return 0;
}

Could you please give me some hints? All my submissions get the same answer "time limit exceeded".

What does it mean?

Есть ли скрипт, позволяющий сравнивать пользователей в виде таблицы по задачам с "+"("АС") - решена, "-"(WA5, TL12 ...) - не решена, " " - не пробовал? Или что-то в этом духе? Простите, за мои плохие знания английского языка.
Is there a script that allows you to compare users in the form of a table for tasks with "+" ("as") - solved, "- "(WA5, TL12 ...) - not solved, "" - not tried? Or something like that? Sorry for my poor English skills.

Edited by author 04.06.2009 17:47

Edited by author 04.06.2009 17:47

It can help you:
Test:
1 13
0 -10 -7 -10 1 -5 6 -10 -8 -8 -7 -10 1
Answer: 6

Edited by author 04.04.2018 15:40

Use printf("%.0lf\n",  floor(answer + .5)); for output answer ;)

I don’t understand what is the mistake here. Help pls
              string[] val = Console.ReadLine().Split(' ');

               double[] k = new double[10000];

              for(int i = 0; i < Convert.ToInt32(val[1]); i++)
               {
                   string s = Console.ReadLine();
                   k[Convert.ToInt32(s) - 1]++;
               }
               double swap = 0;
               for(int i = 0; i < k.Length-1; i++)
               {
                   for(int j = i + 1; j < k.Length - i - 1; j++)
                   {
                       if (k[j+1] > k[j])
                       {
                           swap = k[j+1];
                           k[j+1] = k[j];
                           k[j] = swap;
                       }
                   }
               }
               for (int i = 0; i < Convert.ToInt32(val[0]); i++)
               {

                   Console.WriteLine(string.Format(CultureInfo.InvariantCulture, "{0:N2}", (100 * k[i]) / Convert.ToDouble(val[1]))+"%");
               }

In python 3.6 extra print() in the end of the program gives you WA2. Therefore you should output
"8
 1 3 4 2 0 1 3 2"
instead of
"8
 1 3 4 2 0 1 3 2
 "

a = int(input())
b = int(input())
print(a+b)

Try this:
#X#
#OO
XOX

15 17
1 2
2 3
3 4
4 5
5 6
6 14
14 15
15 9
9 7
7 5
7 8
8 13
13 12
11 12
10 11
10 4
1 6
1 12 15
ans: 3

#include<bits/stdc++.h>
using namespace std;
void prin ( int i)
{
    int j;
    for ( j =1 ; j <= i ;j++)
    {
        cout << "sin(" << j;
        if ( j == i){
        for (j = 0 ; j<i ; j ++)
            cout <<")";
        break;}
        else {
        if (j%2)
            cout << "-";
        else cout << "+";
        }
    }
}
int main()
{
    int i,j,n;
    cin >> n;
    int k = n;
    for (i = 1; i < n ;i++)
    cout << "(";
    for ( i = 1 ; i <= n ; i++,k--)
    {
        prin(i);
        if (i!= n)
            cout << "+"<<k<<")";
    }
    cout << "+1\n";
    return 0;
}

I spent two hours finding this stupid mistake. Do not be like me, do everything with a margin.

4 4 2 3
1 2
1 3
2 4
3 4

ans: 6

12 16 2 30
1 2
2 3
3 1
4 5
5 6
6 4
7 8
8 9
9 7
10 11
11 12
12 10
1 4
1 7
4 10
7 10

ans: 24

6 8 2 30
1 2
2 3
3 1
4 5
5 6
6 4
1 4
2 4

ans: 12

I checked two accepted solution But They are giving different output in same test case.
AC code 1:  https://paste.ubuntu.com/p/RwtFH2s4xK/
AC code 2:  https://paste.ubuntu.com/p/vmrXrdcJN6/

Tese case :
25 28
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 11
11 12
12 13
13 14
14 15
15 16
16 17
17 18
18 1
5 19
19 20
20 21
21 22
22 23
23 24
24 25
25 14
15 22
6 22
1 10 21

Output in code 1 : 2
Output in code 2 : 3

Answer should be : 2