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;
}

It says, that K is an amount of stones BETWEEN other stones. Why does Test4 answer equal n-1. I suppose it should be n-2, shouldn't it?

Edited by author 26.12.2016 16:46

0) AC solve is DP on tree.
1) In DP inf = 1e9.
2) All test in the forum:

5 5
3 1 1
4 1 10
2 3 20
3 5 20
ans = 21

7 4
1 2 10
1 3 20
2 4 30
2 5 0
3 6 0
3 7 40
ans = 100

4 1
1 2 1
1 3 2
2 4 3
ans = 2

5 1
1 2 1
1 5 0
2 3 0
2 4 0
ans = 1

7 3
1 2 1
1 3 39
2 4 40
2 5 40
3 6 39
3 7 39
ans = 117

5 2
1 3 1
1 4 10
2 3 20
3 5 20
ans = 21

Edited by author 12.05.2008 19:54

no matter who plays first and which sign is between numbers

5
5 4 3 2 1
(2)

1
1
(1)

10
3 3 4 2 5 6 1 9 9 3
(4)

you should output edges in input order

a,b = map(int,input().split())
if (a % 2 == 0 and b % 2 != 0):
    print("yes")
else:
    print("no")

Check that you use only shortest routes.
I don't have any tests, but when I started checking this, then I got ACC.

Таким образом, если для данного перекрёстка нет подразделений СКБ Контур находящихся ближе, чем в 6 кварталах, то мы обозначим его числом 0.
Carefully read this part of text.

Try n = 100000 and remember that 1e6 * 1e6 > 2e9.