-> think of doing this problem first -> count of n digits numbers whose sum is equal to x.
->First try to think for smaller numbers less than 1000 i.e from 1 to 999.
-> Create the tree for the same i.e numbers less than 999.
-> after finding f(n,sum)(n is the number of digits and sum is the required sum).
-> you can find f(9,sum)for all numbers less than 1000000000.
-> now finally you need to handle for one number 1000000000.
Cheers

Edited by author 31.07.2020 12:43

I calculated two points of intersection from line that go through two points((x_mid, y_mid) and (x_circle,y_circle)) and then picked nearest two given two points.But this gives me WA3. What should i do?

Is it possible to return my profile to the rating and under what conditions?

Edited by author 04.08.2020 00:40

1.declare an array.size 1000000
2.genarate all number according the problem description.
3.input the value n.
4.declare a container and take the value of the array .vector<int>v(a,a+n);
5.sort the container.
5.at last print v[n];

Edited by author 26.07.2020 04:57

Edited by author 26.07.2020 04:58

If you have TLE on 17-th test , try to use clang compiler. For me it worked))

try this:
3 2
1 3
2 3
1 2

answear: 1

I think this tutorial will be helpful. [video language is Bengali]
https://www.youtube.com/watch?v=MiltLqpzOsE

Edited by author 22.07.2020 19:38

can someone please provide test #6?

check if answer = 2 (Ferma-Euler theorem)

Why i am getting Runtime error (access violation) in test case 9?

Edited by author 21.07.2020 19:22

Edited by author 21.07.2020 19:22

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.