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xxTIMURxx [149]
3 years ago
7

Which is the best Gun is PUBGM1.M4162.SHOTGUN3.CROSSBOW4.GROZAPLS HELP :(​

Computers and Technology
2 answers:
brilliants [131]3 years ago
8 0

Answer:

It all depends on your style and what you prefer or are good at.

Explanation:

If you like mid-long range weapons you might pick an assault rifle or a sniper rifle(Groza, crossbow). If you like a close-mid range weapon you might choose a shotgun or groza.

GrogVix [38]3 years ago
7 0

Answer:

groza

Explanation:

it has the most damage per show with 30 to 40 rounds

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What is application launchers functions​
andrey2020 [161]
An app launcher replaces the stock user interface for organizing the home screen and app icons predominantly in the Android world; however, they are also available for jailbroken iPhones (see iPhone jailbreaking and Cydia). See Launchpad.
5 0
3 years ago
Select the correct statement(s) regarding partial mesh networks.
sweet [91]

Answer: all the answers are correct

Explanation:

Partial mesh networks is simply a packet switching networks which is connected and require that there should be establishment of a circuit before the exchange of packets.

Partial mesh networks are less expensive to implement when they're being compared to the full mesh networks and are also less redundant.

The correct statements regarding partial mesh networks are:

•the number of links connecting nodes in a partial mesh network depends upon the availability and reliability requirements of the network

• switches play an integral role in a partial mesh network, where several paths through the network are available

• redundant links and switches are used to improve the availability and reliability of partial mesh networks

6 0
3 years ago
Flesh out the body of the print_seconds function so that it prints the total amount of seconds given the hours, minutes, and sec
fiasKO [112]

Answer:

Step by step explanation along with code and output is provided below

Explanation:

#include<iostream>

using namespace std;

// print_seconds function that takes three input arguments hours, mints, and seconds. There are 60*60=3600 seconds in one hour and 60 seconds in a minute. Total seconds will be addition of these three  

void print_seconds(int hours, int mints, int seconds)

{

   int total_seconds= hours*3600 + mints*60 + seconds;

   cout<<"Total seconds are: "<<total_seconds<<endl;

}

// test code

// user inputs hours, minutes and seconds and can also leave any of them by  entering 0 that will not effect the program. Then function print_seconds is called to calculate and print the total seconds.

int main()

{

   int h,m,s;

   cout<<"enter hours if any or enter 0"<<endl;

   cin>>h;

   cout<<"enter mints if any or enter 0"<<endl;

   cin>>m;

   cout<<"enter seconds if any or enter 0"<<endl;

   cin>>s;

   print_seconds(h,m,s);

  return 0;

}

Output:

enter hours if any or enter 0

2

enter mints if any or enter 0

25

enter seconds if any or enter 0

10

Total seconds are: 8710

8 0
4 years ago
How many bits are required to encode an image that is 25 pixels wide and 50 pixels tall, if you encode each pixel with 12 bits o
AURORKA [14]

Just multiply all numbers to get the total number of bits:

25*50*12 = 15000 bits.

3 0
3 years ago
a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv
Leokris [45]

Using the knowledge in computational language in C++ it is possible to write a code that  cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

<h3>Writting the code:</h3>

<em>#include <bits/stdc++.h></em>

<em>using namespace std;</em>

<em>// chracter to digit mapping, and the inverse</em>

<em>// (if you want better performance: use array instead of unordered_map)</em>

<em>unordered_map<char, int> c2i;</em>

<em>unordered_map<int, char> i2c;</em>

<em>int ans = 0;</em>

<em>// limit: length of result</em>

<em>int limit = 0;</em>

<em>// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]  </em>

<em>bool helper(vector<string>& words, string& result, int digit, int widx, int sum) { </em>

<em>    if (digit == limit) {</em>

<em>        ans += (sum == 0);</em>

<em>        return sum == 0;</em>

<em>    }</em>

<em>    // if summation at digit position complete, validate it with result[digit].</em>

<em>    if (widx == words.size()) {</em>

<em>        if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {</em>

<em>            if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result</em>

<em>                return false;</em>

<em>            c2i[result[digit]] = sum % 10;</em>

<em>            i2c[sum%10] = result[digit];</em>

<em>            bool tmp = helper(words, result, digit+1, 0, sum/10);</em>

<em>            c2i.erase(result[digit]);</em>

<em>            i2c.erase(sum%10);</em>

<em>            ans += tmp;</em>

<em>            return tmp;</em>

<em>        } else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){</em>

<em>            if (digit + 1 == limit && 0 == c2i[result[digit]]) {</em>

<em>                return false;</em>

<em>            }</em>

<em>            return helper(words, result, digit+1, 0, sum/10);</em>

<em>        } else {</em>

<em>            return false;</em>

<em>        }</em>

<em>    }</em>

<em>    // if word[widx] length less than digit, ignore and go to next word</em>

<em>    if (digit >= words[widx].length()) {</em>

<em>        return helper(words, result, digit, widx+1, sum);</em>

<em>    }</em>

<em>    // if word[widx][digit] already mapped to a value</em>

<em>    if (c2i.count(words[widx][digit])) {</em>

<em>        if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0) </em>

<em>            return false;</em>

<em>        return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);</em>

<em>    }</em>

<em>    // if word[widx][digit] not mapped to a value yet</em>

<em>    for (int i = 0; i < 10; i++) {</em>

<em>        if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;</em>

<em>        if (i2c.count(i)) continue;</em>

<em>        c2i[words[widx][digit]] = i;</em>

<em>        i2c[i] = words[widx][digit];</em>

<em>        bool tmp = helper(words, result, digit, widx+1, sum+i);</em>

<em>        c2i.erase(words[widx][digit]);</em>

<em>        i2c.erase(i);</em>

<em>    }</em>

<em>    return false;</em>

<em>}</em>

<em>void isSolvable(vector<string>& words, string result) {</em>

<em>    limit = result.length();</em>

<em>    for (auto &w: words) </em>

<em>        if (w.length() > limit) </em>

<em>            return;</em>

<em>    for (auto&w:words) </em>

<em>        reverse(w.begin(), w.end());</em>

<em>    reverse(result.begin(), result.end());</em>

<em>    int aa = helper(words, result, 0, 0, 0);</em>

<em>}</em>

<em />

<em>int main()</em>

<em>{</em>

<em>    ans = 0;</em>

<em>    vector<string> words={"GREEN" , "BLUE"} ;</em>

<em>    string result = "BLACK";</em>

<em>    isSolvable(words, result);</em>

<em>    cout << ans << "\n";</em>

<em>    return 0;</em>

<em>}</em>

See more about C++ code at brainly.com/question/19705654

#SPJ1

3 0
2 years ago
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