and also
N=6,K=10
N=7,K=10
N=6,K=2

For some reason, my code gets the right answer for all small inputs (<12) but is 1 greater than both the large test cases I've found.
It gets that s(212) = 995645336, as compared to 995645335, and s(500) = 732986521245024, as compared to 732986521245023 given before.
Anyone have any test cases with an input between 11 and 212 to help me pinpoint this problem?

Edited by author 18.08.2020 20:03

I don't know any string of length 7 contains less than 7 distinct palindrome substrings. Anyone can put the answer for 7, pls?

1)
in:
3 5
..*..
!....
....$
out: 3

2)
in:
5 8
....AA.#
.######!
...BA*##
.#######
..B$...B
out: 3

3)
in:
5 8
....AA.#
.######!
....#*##
.#######
..B$...B
out: Impossible

4)
in:
1 3
$*!
out: Impossible

5)
in:
1 3
*!$
out: 2


I have used BFS, but still WA 22 :-(

Edited by author 26.10.2008 13:57

Edited by author 26.10.2008 14:04

Edited by author 26.10.2008 14:17

Try this test:
3 3
##$
#*#
!##
Answer: Impossible

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main()
{
    int Num = 0, ans = -2;

    cin >> Num;

    vector <int> fact;

    int n = 2;

    while (Num > 1)
    {
        if (n > 9) {
            ans = -1;
            break;
        }
        while (Num % n == 0) {
            Num /= n;
            fact.push_back(n);
        }
        n++;
    }

    sort(fact.begin(), fact.end());

    if (n < 10 & ans != -1)
    {
        ans = 0;
        for (int i = 0; i < fact.size(); i++) {
            //ans += fact.at(i) * (i*10);
            cout << fact[i];
        }
    }


    //cout << ans;
}

Hey! I solved this problem and here are some test that you may check for you solution:

0->10
1->1
2->2
3->3
4->4
5->5
6->6
7->7
8->8
9->9
10->25
11->-1
12->26
13->-1
14->27
15->35
16->28
17->-1
18->29
19->-1
20->45
21->37
22->-1
23->-1
24->38
25->55
26->-1
27->39
28->47
29->-1
30->56
31->-1
32->48
33->-1
34->-1
35->57
36->49
37->-1
38->-1
39->-1
40->58
41->-1
42->67
43->-1
44->-1
45->59
46->-1
47->-1
48->68
49->77
50->255
51->-1
52->-1
53->-1
54->69
55->-1
56->78
57->-1
58->-1
59->-1
60->256
61->-1
62->-1
63->79
64->88
65->-1
66->-1
67->-1
68->-1
69->-1
70->257
71->-1
72->89
73->-1
74->-1
75->355
76->-1
77->-1
78->-1
79->-1
80->258
81->99
82->-1
83->-1
84->267
85->-1
86->-1
87->-1
88->-1
89->-1
90->259
91->-1
92->-1
93->-1
94->-1
95->-1
96->268
97->-1
98->277
99->-1
100->455

Good Luck!

Edited by author 17.08.2020 15:23

Edited by author 17.08.2020 15:23

Edited by author 17.08.2020 15:23

Max distance may be equal 5e4 * 1e3 = 5e7.
I use INF = 1e6 and spent all day to find this bug.

//We need to sort out 3 in the second power of the options for the arrangement of signs

#include <iostream>

using namespace std;

int main()
{
    int a, b, c, min;
    cin >> a >> b >> c;
    min = a + b + c;
    if(a + b - c < min) min = a + b - c;
    if(a - b + c < min) min = a - b + c;
    if(a - b - c < min) min = a - b - c;
    if(a - b + c < min) min = a - b + c;
    if(a - b * c < min) min = a - b * c;
    if(a * b * c < min) min = a * b * c;
    if(a * b + c < min) min = a * b + c;
    if(a * b - c < min) min = a * b - c;
    cout << min;
    return 0;
}

Is it right that p is divisor of n? If not, Can you write example?

I think, this should be renamed to "Cipher Message 4"

Edited by author 13.08.2020 12:24

Edited by author 13.08.2020 12:24

UPD: i have find mistake

Edited by author 13.08.2020 00:16

PLEASE!!!!

Input in the task looks like:
13 14.1

Not like:
13
14.1

So change input in your program to
p,q=map(float, input().split())

0.046 - O(p) solution.
0.468 - O(p^2) solution, please add more tests.
UPD: understood, that there is cant be NO SOLUTION. But you can enlarge p.

Edited by author 10.08.2020 12:39

this is my code
a=int(input())
b = [int(a) for a in input().split()]
c=int(input())
d = [int(c) for c in input().split()]
e=int(input())
f=[int(e) for e in input().split()]
spisok=b+d+f
spisok.sort()
kolich=0
for i in range(0, len(spisok)-1):
    if spisok[i-1]==spisok[i] and spisok[i]==spisok[i+1]:
        kolich+=1
print(kolich)

i dont know why but it writes wrong answer on test №37
please could you text me what is on this test

add ? to your list

#include <bits/stdc++.h>
#include<cstdio>
#include <queue>
#define pb push_back
#define mp make_pair
//Macro
#define eps 1e-9
#define pi acos(-1.0)
#define ff first
#define ss second
#define re return
#define QI queue
#define SI stack
#define SZ(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define sq(a) ((a)*(a))
#define distance(a,b) (sq(a.x-b.x) + sq(a.y-b.y))
#define iseq(a,b) (fabs(a-b)<eps)
#define eq(a,b) iseq(a,b)
#define ms(a,b) memset((a),(b),sizeof(a))
#define G() getchar()
#define MAX3(a,b,c) max(a,max(b,c))
#define II ( { int a ; read(a) ; a; } )
#define LL ( { Long a ; read(a) ; a; } )
#define DD ({double a; scanf("%lf", &a); a;})

double const EPS=3e-8;
using namespace std;

#define FI freopen ("input_B.txt", "r", stdin)
#define FO freopen ("output_B.txt", "w", stdout)

typedef long long Long;
typedef long long int64;


//For loop

#define forab(i, a, b) for (__typeof (b) i = (a) ; i <= b ; ++i)
#define rep(i, n) forab (i, 0, (n) - 1)
#define For(i, n) forab (i, 1, n)
#define rofba(i, a, b) for (__typeof (b)i = (b) ; i >= a ; --i)
#define per(i, n) rofba (i, 0, (n) - 1)
#define rof(i, n) rofba (i, 1, n)
#define forstl(i, s) for (__typeof ((s).end ()) i = (s).begin (); i != (s).end (); ++i)
#define for1(i, a, b) for(int i=a; i<b; i++)

template< class T > T gcd(T a, T b) { return (b != 0 ? gcd(b, a%b) : a); }
template< class T > T lcm(T a, T b) { return (a / gcd(a, b) * b); }
#define __(args...) {dbg,args; cerr<<endl;}

#define __1D(a,n) rep(i,n) { if(i) printf(" ") ; cout << a[i] ; }
#define __2D(a,r,c,f) forab(i,f,r-!f){forab(j,f,c-!f){if(j!=f)printf(" ");cout<<a[i][j];}cout<<endl;}

signed main ()
{
    ios_base :: sync_with_stdio(0);cin.tie(0);cin.tie(0);

    int n;
    unsigned int a;
    priority_queue <unsigned int> pq;

    scanf("%d", &n);

    bool odd = (n%2==1);

    for(int i=n/2; i>=0; --i){
        scanf("%u", &a);
        pq.push(a);
    }
    n-=n/2+1;

    for(int i=0; i<n; i++){
        scanf("%u", &a);
        pq.push(a);
        pq.pop();
    }
    if(odd)printf("%u\n", pq.top());
    else{
        a=pq.top();
        pq.pop();

       // cout << fixed << setprecision(1) << (a+pq.top())/2.0 << endl;
       printf("%.1f\n", (a + pq.top() ) / 2.0);
    }
    return 0;
}