SQL Server – way to find when the system state file is overrun

Is there a way to find the events when moving advanced health event files instead of manually monitoring the events?

For my medium-load server they stay up to 2-3 days. However, on busy servers, these files are updated every 15 minutes, but without fixed patterns or timelines. We know the reason and work to filter out unwanted or reported issues.

I'm curious how we can query when the roll-over of files will take place. I do not see much documentation on MS documents but can not find this information?

Please suggest if and how this is possible.

To get a value from a check box in jquery that is in a column, use .map .find

To get a value from a check box in jquery that is in a column, use .map .find. I am using this code and it does not chain the value of the check box.

                                                                var cols_id_servico = $ (& # 39; col_id_servico & # 39;
.closest ("tr")
.map (function () {
return $ (this) .find ("input: eq (0)") val () + "/" +
$ (this) .find ("input: eq (2)"). val () + "/" +
$ (this) .find ("input: eq (3)"). val () + "/" +
$ (this) .find ("input[type='checkbox']: checked: Eq. (4) "). Length> 0?" Yes No ";
.toArray ()
.join (& # 39;, & # 39;);
alert (cols_id_servico);

Javascript – Find the smallest index that is the same as the value in an array

The task is taken from LeetCode

Given is an array A of different integers sorted in ascending order.
return the smallest index i that satisfies A[i] == i. Returns -1 if no
that's how I exist.

Example 1:

Entrance: [-10,-5,0,3,7]
Edition: 3
// For the specified array, A is[0] = -10, A[1] = -5, A[2] = 0, A[3] = 3, so the output is 3.

Example 2:

Entrance: [0,2,5,8,17]
Output: 0
// ON[0] = 0, so the output is 0.

Example 3:

Entrance: [-10,-5,3,4,7,9]
Output: -1
// There is no such i, that A[i] = i, so the output is -1.


1 <= A. Length <10 ^ 4

-10 ^ 9 <= A[i] <= 10 ^ 9

My solution

has temporal complexity of $ O (n) $ and space complexity of $ O (1) $, I start to look from the beginning to the last element. If I find a value that is greater than I, then I can leave early (because there is no element that is the same I no more). If I find … ON[i] === ithen I have a result.

Is there a faster solution than the intended one?

/ **
* @param {number[]} ON
* @return {number}
* /
var fixedPoint = function (A) {
for (let i = 0; i < A.length; i++) {
        if (A[i] > i) {return -1; }
when a[i] === i) {return i; }
return -1;

Separate Witness – Find the sender's public key in SegWit Transaction

The public key you mentioned is indeed the sender's key and was used to derive the P2SH address (P2PKH): 35yfMa3CRBiWny8DFdb4tUu9fn7fcdvVp9, The way in which a P2SH address (P2PKH) is derived is as follows:

1. witness_script = hash160 (pub_key) #this is equal to & # 39; 4b9d2d3dd1174ad656754a0c664e7a129b131f3b & # 39;
2. witness_version = 0x00 # current SegWit version
3. scriptSig = witness + OP_DATA (0x14) + witness_script
#Above is equal to & # 39; 00144b9d2d3dd1174ad656754a0c664e7a129b131f3b & # 39; and is displayed in scriptSig in Explorer.
# 0x14 instructs the script to write the next 20 bytes to the stack
4. public_address = hash160 (scriptSig)
5. bitcoin_address = base58check (public_address) # with the prefix 0x05
Result = 35yfMa3CRBiWny8DFdb4tUu9fn7fcdvVp9

Native SegWit transactions work by locking an output scriptPubKey: version + OP_DATA (bytes to transfer) + witness_script, For older customers, it looks like everyone could issue a transaction because there is no opcode that confirms anything. To use the lower fees for SegWit transactions while being compatible with wallet software that SegWit does not know, we use the scriptPubKey as the script and create a P2SH lock script out of it. The lock script is thus: OP_HASH160 OP_EQUAL, (The public_address convention is the same as the one I mentioned while deriving the above address.) A customer can send you bitcoins without updating their wallet software to be SegWit-enabled. However, if you spend the bitcoins, you can use the lower fees of SegWit.

Now to check when this issue is issued in a transaction: Older clients checking the transaction look at only those scriptSig, take his has160, check with that public address and consider this transaction as valid. Newer customers will find that the scriptSig is in itself a SegWit and looks in the witness part of the transaction for signatures. There, the clients make sure that the hash160 of the public key matches the witness_script, and the signature then matches the public key if it is signed with the transaction as a message. The signature message digest is described in BIP 143.

Calculation and Analysis – Can anyone suggest a reference where I can find out more about the function of Conway Base 13 in more detail

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Transactions – Find the sender's public key in the Segwit block

How can I get the sender's public key in a Segwit transaction?

Example: https://www.blockchain.com/btc/tx/3179d18d6f36fa77b88909496535485d8188d96b79d295843789a9e0ff6c3e6a

I like to analyze the witness share as in

02: 2 articles follow

47: 0x47 = 71 bytes (signature)


21: 0x21 = 33 bytes (the pubkey)

02384052a5ecde83bf8ee7ed77f378edb58aa65de22c4e91af87eee68015b9d509: the actual pubkey

But this pubkey is Not the public key of the sender. It's probably the pubkey of a Segwit (or something like that) …

How exactly is this signature composed and which pubkey is that? And more importantly, how can I get the sender's public key (which in this example belongs to 35yfMa3CRBiWny8DFdb4tUu9fn7fcdvVp9)?

Thanks for all the hints

Find c ++ prime. with strainer of Eratosthenes

Link from Promble
I try to solve this Projecteuler question 7, but get an incorrect answer.

right answer in "104743" but I get "104759" Please help to find what is wrong.

                // with Sieve of Eratosthenes


using namespace std;

vector to {};

int cprime (long long int n) {
vector array;
int f = 0;

for (int i = 0; i <= n + 1; i ++) {
array.push_back (1);

array[0]= 0;
array[1]= 0;

for (int i = 2; i <= sqrt (n); i ++) {
if (array[i]== 1) {
for (int j = 2; i * j <n; j ++) {
array[i*j]= 0;

for (int i = 2; i <n; i ++) {
if (f == 10001 && array[i]== 1) {
cout << i << endl;
if (array[i]== 1) {
f ++;


int main () {

Cprime (1000000);

return 0;