For everyone got TLE (the time limits are very strict).

I implemented the dijkstra by priority_queue and got TLE with testcase 7 and 8. After that, I added an array visited[] to ...

not able to understand the can someone make me understand why dijkstra is use

Thanks a lot for the link!

If you are failing case 15. Reread the question several times. I just misunderstood the question otherwise i would have gotten correct in one try.

Got AC

#include <iostream>
#include <cstring>
 
using namespace std;
int N;

/* package codechef; // don't place package name! */
 
import java.util.*;
import java.lang.*;
import java.io.*;

package java102;

public class simplecalc {

public static void main(String[] args) {

t = int(input())
a,b=0,0
prevp1,prevp2 = 0,0
while(t>0):
    l = list(map(int,input().split()))

#include<bits/stdc++.h>

using namespace std;

WA for 7 test cases. I tried it without k -= k > 0 ? 1 : 0 but it always seems that I have to decrease k by 1 to get the right answer. Please help!

Here is my solution: Solution: 28918761

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int ansf[1000005];

Actually BIT supports max/min queries and continuous sum( a[i] ~ a[i+L] ==> +x) queries.

We can implement by using 2 BIT.

Oops, should have reread my reply before posting, thanks for the correction.

'Almost every solution you can do with Segment Tree can also be done with Binary Index Trees.' Its true the other way round because BIT only supports sum queries

def bs(start,end,key):
  while(start<end):
    mid=(start+end)//2
    if key>=arr[mid][0]:
      start=start+1

def binary_search(arr,key,l):
    low=l
    high=len(arr)-1
    while(low<=high):
        mid=(high+low)//2

Can you please explain the reason behind taking the upper limit for binary search n (and not n-1)?

In my case taking upper limit to n-1 gives segmentat...