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!

3 5
1 4 5
7 3 8
2 4 4
7 5 3
7 7 0

3 5
2 5 1
5 3 4
3 8 8
8 6 0
6 6 1
I got tle on these tests.But 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?

In this test N=50 and the answer is 1069.

I spent about an hour till i figured out that:

If it is impossible to get from the computer to the refrigerator, you should output 0 0. !!!!