IF you have TL, try this test:
18
aasda0sd basdasdasd casdasd
asdasd33a 0b12asdasd ca45sdasd
cdasdas3das dasd0asda2ds e1dasqew
sdsdad4asdsaf dasda0sdasd 0gasdasd
asdasda5sa asdasdasd3b asda0sdasdc
aqwewq6e bqweqwe cqweq0we
cqweqw7e dqweqwe eqw3ewqees
fsadasd8asd dasdasdas0d gasdasd
asdasds9adasa abasdas3dads casdasd12
qweqwew123qea bqweqw3ewqe cweqqwee
csadas12d dasdas0d ed0dsdadsa
fdasdasd33 dasdasds30ad gasasdsad
adasdasd123d badas3ddasda cdasds0adas
asdads455as basdasda2sd casdasdd0sa
cqwewq123e deeqw12e eqw1easda8d
fasd123asd dasdasds3a1dqw gqwe9wqewes
jeqweqw123e eqweq123wewee hwewe9wqq
ldszczx123c ad123sdasdaq qweweq09f

ans: 18

Edited by author 10.08.2019 22:28

Edited by author 10.08.2019 22:28

Edited by author 10.08.2019 22:28

#include<bits/stdc++.h>
using namespace std;
int main()
{
    double a,b,c,d,m,n,l,p,t,s,k;
    cin>>a>>b>>c>>d>>m>>n>>l;
    p=sqrt((a-m)*(a-m)+(b-n)*(b-n));
    t=sqrt((c-m)*(c-m)+(d-n)*(d-n));
    s=max(p,t);
    if(s>l){s-=l;
    }
    else{ s=0;}
    k=min(p,t);
    if(k<=l){
        k=0;
    }
    else if(p==t){
        a=(a+c)/2,b=(b+d)/2;
        k=sqrt((a-m)*(a-m)+(b-n)*(b-n));
        if(k<=l){
        k=0;
    }
    else{
        k-=l;
    }
    }
    else{
    while(abs(p-t)>.001){
        if(p<t){
            c=(a+c)/2,d=(b+d)/2;
            t=sqrt((c-m)*(c-m)+(d-n)*(d-n));
        }
        else{
            a=(a+c)/2,b=(b+d)/2;
            p=sqrt((a-m)*(a-m)+(b-n)*(b-n));
        }
    }
    k=min(p,t);
        if(k<=l){
        k=0;
    }
    else{
        k-=l;
    }
    }
    cout<<fixed<<setprecision(2)<<k<<endl;
    cout<<fixed<<setprecision(2)<<s<<endl;
}

There seems to be some junk at the end of test 27. My solution with "while (read()) solve();" got WA but when I removed it I got OK.

Методы Программирования и Прикладные Алгоритмы

My solution is O(n) (<=60 * n iterations) but I have TL on python
Help me to read fast data input

My input code:
n, l, r = map(int, input().split())

tokens = sys.stdin.read().split()
a = list(map(int, tokens))

I used some methods, got TLE || WA ...T T ...

You can enumerate P .

Think carefully and you will get a O(sqrt(n)) solution.

#include<bits/stdc++.h>

using namespace std;

int main(){
    unsigned long long int n;
    try{
        cin >> n;
        cout << n%7 << endl;
    }catch(exception e){
        cout << 1 << endl;
    }
    return 0;
}

Some have claimed that this is a very difficult problem. Not true! There is a reasonably simple and very fast DP-solution to this problem. No string algorithms are required beyond the naive ones that every non-programmer knows.

Hint: Store the solution for sentences of length k in A[k], and store the number of strings that end with some forbidden word i (0 < i < P), but does not contain any other forbidden words as substrings in F[i][k]. Then construct the solution from A[k] to A[k+1], updating F[i][k+1] as well for each i. For each new character that you tuck onto the end of a valid string of length k, you get A[k] more valid strings, except those that end with a forbidden word, so those strings need to be removed. But make sure you don't remove stuff that you've assumed that you already removed earlier in the string - that's where F[i][k] comes in handy.

every body handshakes all else his couple
so
n*(n-1) div 2
-
k (couples doesnt handshake eachother)!
-->

Var
  n,k:integer;
begin
  readln(n,k);
  writeln(((n*(n-1)) div 2) -k);
end.

Aidin_n7

This is my solution O(M log N)

#include <bits/stdc++.h>
#define in(x) freopen(x,"r",stdin)
#define out(x) freopen(x,"w",stdout)
#define N 100500
#define oo ll(1e16)
#define P pair < int, int >
#define PP pair < pair < int, int >, int >
#define F first
#define S second
#define pb push_back
#define el endl
#define base ll(1e9 + 7)
using namespace std;
typedef long long ll;
typedef long double ld;
int n, m, y;
bool o;
int bin(int x, int st) {
    int res = 1;
    for(; st > 0;) {
        if (st & 1) res = (res * x) % m;
        x = (x * x) % m;
        st >>= 1;
    }
    return res;
}
int main() {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    cin >> n >> m >> y;
    o = 0;
    for (int i = 1; i < m; i++) {
        int cur = bin(i, n);
        if (cur == y) cout << i << ' ', o = 1;
    }
    if (!o) cout << -1;
}

Use scanline y = -x + a
We need to use no more than 6 points
If you want to be fast use k-statistic in O(n)
For better explanation write here: myironmistake@gmail.com

Edited by author 06.08.2019 18:42

Even though it took me almost 2 hours to solve I'm surprised this has difficulty of 1539. It doesn't require any special knowledge and almost anyone can solve this. Solution is pretty straightforward.

Anyone can give me this test case?

DELETED

Edited by author 04.08.2019 22:55

Dear admin:
  Please add this test:
  3 2
  4 6 2
  1 7 8

  I got tle on this test.But fortunately I got AC!

My two AC submissions give different answers on this test:

3 4
0 0 0
0 6 1
5 5 2
8 5 3

P.S. The first test case isn't the sample.

Edited by author 19.11.2013 01:26

I WAed many many times, and finally got AC. Test6 is 0, whose answer is 2. Test 7 is probably 1 or 2, which my previous program had given incorrect answer to. The answer to Test 8 has to be 36, because I used "for(i=x;i<36;i++)" and wa8. (I also saw some prog posted on this forum had the same problem) When I found this and changed into "<=", AC at last. Sorry for my poor English. Hope these can help you guys somehow. Good luck!

Hi, why is a problem author talking about bubble sort? And why a lot using stl stable_sort? Where is a differents of stl sort?