What is an online alternative to customers sending checks via mail?

A ________ is an application program that runs on a server-tier computer and manages processes such as items in a shopping cart

garik1379 [7]

Commerce Server

An application program that runs on a server tier computer. A typical commerce server obtains product data from a database, manages items in users' shopping carts, and coordinates the checkout process.

6 0

2 years ago

This function receives first_name and last_name, then prints a formatted string of "Name: last_name, first_name" if both names a

pishuonlain [190]

Answer:

Following are the program in the C++ Programming Language.

//set header file

#include <iostream>

//set namespace

using namespace std;

//define class

class format

{

//set access modifier

public:

//set string type variable

string res;

//define function

void names(string first_name, string last_name)

{

//set if-else if condition to check following conditions

if(first_name.length()>0 && last_name.length()>0)

{

res="Name: "+last_name+", "+first_name;

}

else if(first_name.length()>0 and last_name.length()==0)

{

res="Name: "+first_name;

}

else if(first_name.length()==0 and last_name.length()==0)

{

res="";

}

//define function to print result

void out(){

cout<<res<<endl;

}

};

//define main method

int main() {

//set objects of the class

format ob,ob1,ob2;

//call functions through 1st object

ob.names("John","Morris");

ob.out();

//call functions through 2nd object

ob1.names("Jhon","");

ob1.out();

//call functions through 3rd object

ob2.names("", "");

ob2.out();

}

Output:

Name: Morris, John

Name: Jhon

Explanation:

Following are the description of the program:

Define class "format" and inside the class we define two void data type function.

Define void data type function "names()" and pass two string data type arguments in its parameter "first_name" and "last_name" then, set the if-else conditional statement to check that if the variable 'first_name' is greater than 0 and 'last_name' is also greater than 0 then, the string "Name" and the following variables added to the variable "res". Then, set else if to check that if the variable 'first_name' is greater than 0 and 'last_name' is equal to 0 then, the string "Name" and the following variable "first_name" added to the variable "res".
Define void data type function "out()" to print the results of the variable "res".

Finally, we define main method to pass values and call that functions.

3 0

3 years ago

Is ryzen really better for all-around purpose, and intel is better for gaming?

barxatty [35]

Answer:

Its really personal preference G.

Explanation:

But what you said is about right.

6 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>

#include <bits/stdc++.h>

using namespace std;

// chracter to digit mapping, and the inverse

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

unordered_map<char, int> c2i;

unordered_map<int, char> i2c;

int ans = 0;

// limit: length of result

int limit = 0;

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

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

if (digit == limit) {

ans += (sum == 0);

return sum == 0;

}

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

if (widx == words.size()) {

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

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

return false;

c2i[result[digit]] = sum % 10;

i2c[sum%10] = result[digit];

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

c2i.erase(result[digit]);

i2c.erase(sum%10);

ans += tmp;

return tmp;

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

if (digit + 1 == limit && 0 == c2i[result[digit]]) {

return false;

}

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

} else {

return false;

}

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

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

return helper(words, result, digit, widx+1, sum);

}

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

if (c2i.count(words[widx][digit])) {

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

return false;

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

}

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

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

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

if (i2c.count(i)) continue;

c2i[words[widx][digit]] = i;

i2c[i] = words[widx][digit];

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

c2i.erase(words[widx][digit]);

i2c.erase(i);

}

return false;

}

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

limit = result.length();

for (auto &w: words)

if (w.length() > limit)

return;

for (auto&w:words)

reverse(w.begin(), w.end());

reverse(result.begin(), result.end());

int aa = helper(words, result, 0, 0, 0);

}

int main()

{

ans = 0;

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

string result = "BLACK";

isSolvable(words, result);

cout << ans << "\n";

return 0;

}

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

#SPJ1

3 0

2 years ago

How are some businesses capitalizing on social media at the time of someones death

stiv31 [10]

They'll post condolences messages etc which means many people will discover their business if they are looking for posts mentioning the deceased's name on social media.

7 0

3 years ago