who can tell me why it is 670 when the number is 4?

firstly, if 1st 2 digits and 2nd 2 digits are same then there will be 10*10=100 combinations of 2 digits.
for every number there will be 2 combination.(example:1212 and 1221)
so,combinations will be 2*100=200. But there are such 10 numbers for which 2 combinations are not available.(ex: 0000 , 1111 , 5555)
COMBINATIONS = 200-10 = 190.
now, we have to check from highest 9+9=18 to lowest 0+0=0 that how much sets have same sum.
     16<<<< (9+7),(8+8)
     15<<<< (9+6),(8+7)
     14<<<< (9+5),(8+6),(7+7)
     13<<<< (9+4),(8+5),(7+6)
     12<<<< (9+3),(8+4),(7+5),(6+6)
     11<<<< (9+2),(8+3),(7+4),(6+5)
     10<<<< (9+1),(8+2),(7+3),(6+4),(5+5)
      9<<<< (9+0),(8+1),(7+2),(6+3),(5+4)
      8<<<< (8+0),(7+1),(6+2),(5+3),(4+4)
      7<<<< (7+0),(6+1),(5+2),(4+3)
      6<<<< (6+0),(5+1),(4+2),(3+3)
      5<<<< (5+0),(4+1),(3+2)
      4<<<< (4+0),(3+1),(2+2)
      3<<<< (3+0),(2+1)
      2<<<< (2+0),(1+1)

  combinations =(2*2*2/2)+(2*2*2)+(2*2*2+2*2+2*2) +(3*2*2*2)+..................................+(2*2*2)+(2*2) =480

SO, total combinations = 190+480 =670

if (nuts / 2 % 2 == 0)
    {
        daenerys_wins = !daenerys_wins;
}

print "IMPOSSIBLE" instead of "Impossible"

in the question , author said 0 is replaced by 1.
but they replace 1 by zero in the 3rd test of sample.
now, i am very much confused.

gimme test, please)

var
n:integer;
s:real;
begin
read(n);
s:=(n+2)*((n+1)*n)/2;
write(s);
end.

Edited by author 23.06.2019 11:35

#include<iostream>
using namespace std;
void sin (int n,int i)
{
    if(i==n+1)
      return ;
    else
    {
      if(i==1)
        {
          cout<<"sin("<<i;
          sin(n,i+1);
        }
      else if(i%2==1)
        {
          cout<<"+sin("<<i;
          sin(n,i+1);
        }
      else
        {
          cout<<"-sin("<<i;
          sin(n,i+1);
        }
       cout<<")";
    }

}
void sol( int n,int i)
{
  if(i==n)
    return;
  if(i==1)
    for(int i=1;i<n;i++)
      cout<<'(';
  sin(i,1);
  cout<<'+'<<n-i+1<<')';
  sol(n,i+1);

  if(i==1)
  {
    sin(n,1);
    cout<<"+1";
  }
}
int main ()
{
  int n;
  cin>>n;
  sol(n,1);
}

Can somebody tell me how solve this problem?

It may be caused by conflicts...
(Sorry for my poor English

what is test 4. plz help me.
my code is below
#include<bits/stdc++.h>
using namespace std;
int main()
{
    long n=1000,i,k;
    long long a[100000];
    a[0]=0,a[1]=1;
    for(i=2;i<n;i=i+2){
            k=i/2;
        a[i]=a[k],a[i+1]=a[k]+a[k+1];
    }
    cin>>n;
    while(n){
            cout<<*max_element(a,a+n+1)<<endl;
        cin>>n;
    }
}

Edited by author 22.06.2019 13:46

#include <iostream>
int n,x;
int a[10];
int i,t;

int _max(int s1,int s2,int x)
{
    if (x==n)
        return s1+s2;
    else {
        int t1,t2;
        if (x*2-1<=n)
            t1 = _max(s1,s2+s1,x*2-1);
        else
            t1 = 0;
        if (x*2+1<=n)
            t2 = _max(s1+s2,s2,x*2+1);
        else
            t2 = 0;
        if ((t1==t2)&&(t2==0))
            return s1+s2;
        else
            return t1>t2?t1:t2;
    }
}

int main()
{
    std::cin >> n;
    i = 0;
    while (n){
        if (n==2)
            a[i] = 1;
        else
        if (n==1)
            a[i] = 1;
        else
        if (n==0)
            a[i] = 0;
        else
            a[i] = _max(1,1,3);
        std::cin >> n;
        i++;
    }
    for (n = 0;n<i;n++)
        std::cout << a[n] << "\n";

    return 0;
}

Since my solution got AC, there IS a limit on k... or the tests are too weak.

Hi! I'm just amazed about this algorithm! My previous O(N^3) implementation take 0.624 and 508 KB with my best efforts to reduce runtime. When I submitted the Manacher's one, the runtime dropped down to 0.015 and 2376 KB!! Just google it. It's a very easy and efficient O(N) algorithm. That's all you need to solve this problem. No more than 30 lines of code.

Can anyone please tell me in tabular form what is are the values of answers for the above values. My algorithm is almost right and it is calculating accurately as far as I know but it is always responding No on test # 2. Please help!!! I am stuck and dissappointed.... :(

What is the test 18? I really don't understand it... I wrote solution with ternary search, and I think that there is problem with accuracy of calculations. Have anybody this problem?

P.S. Sorry for my bad english :)

The following code is failing the test case 3, I have used linked-list to store the data in as a stack. Below is the code for reference, can you tell me where I am going wrong. Thank you.

#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
#include<math.h>
#include<assert.h>

typedef unsigned long long _ull;
typedef unsigned int _uint;

typedef struct _link_list {
    _ull _number;
    struct _link_list*_next;
} _list;

void _makenode(_list**,_ull);
void _insertnode(_list**,_list*);

int main(int argc,const char*argv[]) {
    _ull _val;
    _list *_head,*_temp;
    _head = _temp = NULL;
    while(true) {
        fscanf(stdin,"%lld",&_val);
        assert(_val>=0);
        if(feof(stdin)) break;
        _makenode(&_head,_val);
    }
    while(_head) {
        _temp = _head;
        fprintf(stdout,"%0.4lf\n",sqrt(_head->_number));
        _head = (_head->_next);
        free(_temp);
    }
    return 0;
}

void _makenode(_list**_ptr,_ull _data) {
    _list*_node = malloc(sizeof(_list));
    (_node->_number) = _data;
    (_node->_next) = NULL;
    _insertnode(_ptr,_node);
}

void _insertnode(_list**_ptr,_list*_node) {
    if(!(*_ptr)) (*_ptr) = _node;
    else {
        (_node->_next) = (*_ptr);
        (*_ptr) = _node;
    }
}

Try testcases
12 11
1
12 10
2
100 7
7
20 7
2
1000 1
26
10000 1
37
10000 13
25
100 25
8