Consider the following variation on the change-making problem: you are given denominations x1; x2; : : : ; xn, and you want to make change for a value v, but you are allowed to use each denomination at most once. For instance, if the denominations are 1; 5; 10; 20, then you can make change for 16 = 1 + 15 and for 31 = 1 + 10 + 20 but not for 40 (because you can't use 20 twice).

answer:

explanation:

answer:

Ask a question and ill answer

Explanation:

answer:

// here is code in C++.

#include <bits/stdc++.h>

using namespace std;

//main function

int main()

{

  // variable

 int n;

 cout<<"enter a number:";

 // read the input

 cin>>n;

 // find the square root of the absolute value

 float square_root=sqrt(abs(n));

 // print the square_root

 cout<<"square root of "<<n<<" is: "<<square_root<<endl;

  return 0;

}

Explanation:

Read a number from user and assign it to "n".Then find its absolute value with abs() function.Then calculate the square root of the absolute value with sqrt() function.Print the square root of the number.

Output:

enter a number:-17

square root of -17 is: 4.12311

                                                         This is easy

                                                     sinserly Natalie 7 grader (this is not that hard)