OJ Board

Jehad Uddin

O(n^2 log n) is enough for the problem.Take two point and find the other two . Then check two point from the map in log n.

Jehad Uddin

@kbr_iut
iterative dp is faster with same approach.
run time reduced to 0.256 from 1.088 sec.

Jehad Uddin

hello ghana,

change ur pointer calculation

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char bin[50];
void calculate(int n)
{

use code tags while posting code

Jehad Uddin

your implementation of tarjans algorithm is not correct. see http://en.wikipedia.org/wiki/Tarjan's_strongly_connected_components_algorithm there are some pop operations you are missing int pp; void tarjan( int x ){ ind[ x ] = c; low[ x ] = c; c ++; node.push( x ); for( int i = 1; i <= N; i ++ ){ if(...

Jehad Uddin

use long long to avoid wa

Jehad Uddin

219760

Jehad Uddin

thanks

Jehad Uddin

getting wa pls help

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got ac code removed :)

Jehad Uddin

you can precalculate on ans

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for(i=1;i<21;i++)
 {
        ar[i][1]=1;
        ar[1][i]=i;
 }

    for(i=2;i<21;i++)
    {
        for(j=2;j<=55;j++)
        {
             ar[i][j]=1+ar[i-1][j-1]+ar[i][j-1];
       }
    }

then just chek ar[s][k]>=N then output k

Jehad Uddin

my code passes all the cases of forum , but still i am getting wa ,pls help :oops: :oops: #include<stdio.h> #include<iostream> using namespace std; long long dp[80][80][10]; long long grid[80][80]; long long b[80][10]; long long c[80][10]; int n,k; int main() { int tst=0,i,j,l; long long INF=9000000...

Jehad Uddin

dp dimension is [length][stateOfNo][lastBit][AnyAdjacentOne] ,it will return how many nos having 11 pattern,
stateOfNo is whether it is <=N
AnyAdjacentOne is whether 11 occurs or not
Using this you can calculate number of 11s using another 4d BigIntArray

Jehad Uddin

@ Piotr42
you can solve it using dp
dimension is [n][k];

Jehad Uddin

i am getting tle please help

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code removed got acc after 16 submission

Jehad Uddin

thank you

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code removed got acc.

Jehad Uddin

i got wa in this code,but when i switched to java ,in same process i got ac
this is the wa code

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code removed...

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