In SI unit, velocity is measured in meters per second. This unit doesn't have a special name.
The formular is:m/s
Princewill Esara
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∙ 14y agoA study area.
velocity = velocity
muzzle velocity is the velocity of bullet and recoil velocity is the velocity of gun.
That would be velocity
The answer is velocity.
Liter
The number of digits is immaterial, the principal is exactly the same. The key to reversing any number is in the following user-defined function: uint reverse(uint num) { uint rev=0; while(num) { rev*=10; rev+=num%10; num/=10; } return(rev); } The following code demonstrates how to reverse a 6 digit number entered from the console. #include<iostream> #include<string> #include<sstream> typedef unsigned int uint; uint reverse(uint num) { uint rev=0; while(num) { rev*=10; rev+=num%10; num/=10; } return(rev); } int getrange(uint digits,uint& min,uint&max) { if(!digits) return(-1); min=1; while(--digits) min*=10; max=min*10-1; return(0); } uint inputnumber(const std::string prompt,const uint min,const uint max) { using namespace std; uint result=0; while(1) { cout<<prompt; string s; getline(cin,s); if(s.size()) result=stoul(s,0,10); if(result<min result>max) cout<<"The number must be in the range " <<min<<".."<<max<<"\n" <<"Please enter another number.\n"<<endl; else break; } return(result); } uint inputnumber(const uint digitcount) { using namespace std; uint min=0, max=0; if(getrange(digitcount,min,max)==-1) return(0); std::stringstream prompt; prompt<<"Enter a "<<digitcount<<" digit number: "; return(inputnumber(prompt.str(),min,max)); } int main() { using namespace std; uint num=inputnumber(6); // 6 digits. if(num) { uint rev=reverse(num); cout<<"Reverse: "<<rev<<"\n"<<endl; } return(0); } Output: Enter a 6 digit number: 123456 Reverse: 654321
You mean an iterative function, not a recursive one. A recursive function is one that calls itself with altered arguments, typically used to implement divide-and-conquer algorithms. Since the arguments remain the same and you're not dividing the array into smaller subsets, recursion is not necessary. An iterative loop is all you need: typedef unsigned int UINT; const UINT CountOddNumbers( const UINT* const A, const UINT size) { UINT result=0; for(UINT i=0; i<size; ++i) result += A[i]&1; return(result); } Note that this function is not safe, since size is not guaranteed to reflect the actual size of A. In C++, it's better to use a vector object, since vectors encapsulate the actual size of the embedded array: const UINT CountOddNumbers( const std::vector<UINT> A) { UINT result=0; for(UINT i=0; i<A.size(); ++i) result += A[i]&1; return(result); }
There are many ways to implement Pascal's Triangle in C++. One of the easiest ways is to use a vector of vectors, which is essentially a two-dimensional dynamic array. Although you could use a static 2D array, the matrix would need to be sparse because the first row has only one value, while the second has two values, and so on. Unused elements would need to store the value zero, but unused elements are wasteful. You could improve upon this by creating a 1D array of dynamic arrays, however C++ vectors are much easier to work with and achieve the same thing. The following program asks the user to enter the top value of the triangle and the number of rows in the triangle. The range of acceptable values has been limited to ensure all printed triangles will fit comfortably within an 80-character console window. The program employs a simple class definition to encapsulate the triangle and its methods. The default constructor builds the triangle using the top value and the number of rows supplied from the user-input, while the print function displays the triangle's values in a symmetric triangle formation. #include<iostream> #include<iomanip> #include<vector> #include<string> #include<sstream> // Some typdefs to simplify coding. typedef unsigned int uint; typedef std::vector<uint> row; // Simple class definition. class pascal_triangle { public: pascal_triangle(uint top=1, uint rows=10); void print(); private: const uint get_width(); std::vector<row> m_triangle; }; // Default constructor: pascal_triangle::pascal_triangle(uint top, uint rows) { for(uint i=0;i<rows;++i) { row r; uint x=top; for(uint k=0;k<=i;++k) { r.push_back(x); x=x*(i-k)/(k+1); } m_triangle.push_back(r); } } // Return the width required for the largest value. const uint pascal_triangle::get_width() { // Reference the last row of the triangle. row& r=m_triangle[m_triangle.size()-1]; // Determine the largest value in that row. uint largest=0; for(uint i=0;i<r.size();++i) if(r[i]>largest) largest=r[i]; // Determine required width plus one space. uint width=1; do { largest/=10; ++width; } while( largest ); // Return an even width. return(width+width%2); } // Print the given triangle: void pascal_triangle::print() { // Call private method to determine width. const uint width=get_width(); // Repeat for each row... for(uint i=0;i<m_triangle.size();++i) { // Insert half-width padding spaces. std::cout<<std::setw((m_triangle.size()-i)*(width/2))<<' '; // Print the row values. row& r=m_triangle[i]; for(uint j=0;j<r.size();++j) std::cout<<std::setw(width)<<r[j]; std::cout<<std::endl; } std::cout<<std::endl; } // Print the given range: void print_range(const uint min,const uint max) { std::cout<<'['<<min<<".."<<max<<']'; } // Return a natural number in the given range from user input: uint enter_natural(const std::string& prompt,const uint min,const uint max) { uint result=0; while(!result) { std::cout<<'\n'<<prompt<<' '; print_range(min,max); std::cout<<": "; std::string in; std::getline(std::cin,in); std::stringstream(in)>>result; if(result<minresult>max ) { std::cout<<"The valid range is "; print_range(min,max); std::cout<<"\nPlease try again.\n"<<std::endl; result=0; } } std::cout<<std::endl; return(result); } int main() { std::cout<<"Pascal's Triangle\n"<<std::endl; uint top=enter_natural("Enter top value",1,10); uint rows=enter_natural("Enter number of rows",2,10); pascal_triangle t(top,rows); t.print(); return(0); }
A study area.
Matrix multiplication comes in many forms. In its simplest form, both matrices are the same size, and you simply multiply the corresponding elements in each matrix to produce a third matrix of the same size, known as the matrix product. The following code demonstrates this form only. Note that the code uses a vector of vectors to simulate a 2 dimensional array of unsigned int and encapsulates the array within a class. This allows you to not only generate dynamic arrays more easily, but also allows you to implement the matrix product as a multiplication operator overload. The code demonstrates generating a new matrix product array from two existing arrays (a3=a1*a2) as well as mutating an existing array into a matrix product from another array (a1 *= a2). However, the code will not handle arrays of different size or orientation. #include<iomanip> #include<iostream> #include<time.h> #include<vector> typedef unsigned int uint; typedef std::vector<uint> col_array; typedef std::vector<uint>::iterator col_it; typedef std::vector<col_array> row_array; typedef std::vector<col_array>::iterator row_it; class cArray { public: cArray(uint rows, uint cols):m_rows(rows),m_cols(cols){} cArray operator* (const cArray& rhs); cArray& operator*= (const cArray& rhs); void generate(); void print(); std::string name; private: uint m_rows; uint m_cols; row_array m_data; }; void cArray::generate() { m_data.clear(); for(uint r=0; r<m_rows; ++r) { col_array col; for(uint c=0; c<m_cols; ++c) col.push_back( rand()%10+1 ); m_data.push_back( col ); } } void cArray::print() { std::cout << "Array name: " << name.c_str() << "\n" << std::endl; for(uint r=0; r<m_rows; ++r) { for(uint c=0; c<m_cols; ++c) std::cout << std::setw(4) << m_data[r][c]; std::cout << std::endl; } std::cout << std::endl; } cArray cArray::operator* (const cArray& rhs) { cArray result(m_rows,m_cols); result.name = name; result.name += " * "; result.name += rhs.name; for(uint r=0; r<m_rows; ++r) { col_array col; for(uint c=0; c<m_cols; ++c) col.push_back(m_data[r][c] * rhs.m_data[r][c]); result.m_data.push_back(col); } return( result ); } cArray& cArray::operator*= (const cArray& rhs) { for(uint r=0; r<m_rows; ++r) for(uint c=0; c<m_cols; ++c) m_data[r][c] *= rhs.m_data[r][c]; return( *this ); } int main() { srand((uint)time(NULL)); uint rows = (uint) rand()%9 + 2; uint cols = (uint) rand()%9 + 2; std::cout << "Rows = " << rows << "\n"; std::cout << "Cols = " << cols << "\n" << std::endl; cArray a1(rows, cols); cArray a2(rows, cols); a1.name="Left"; a1.generate(); a1.print(); a2.name="Right"; a2.generate(); a2.print(); cArray a3 = a1 * a2; a3.print(); std::cout << "Compound assignment (" << a1.name.c_str(); std::cout << " *= " << a2.name.c_str() << ")\n" << std::endl; a1 *= a2; a1.print(); return(0); }
The following code demonstrates one way to convert from binary to octal based on user input: #include<iostream> #include<string> typedef unsigned int uint; std::string input(std::string prompt) { std::cout<<prompt<<": "; std::string s; std::getline(std::cin,s); return(s); } uint input_binary( uint min, uint max ) { uint result=0; bool repeat=true; while(repeat) { result=0; repeat=!repeat; std::string s=input("Enter a binary value"); for(uint i=0; i<s.size() && !repeat; ++i) { const char c=s.at(i); if(c<'0'c>'1') { std::cout<<"Binary values may only contain characters '0' or '1'\n"<<std::endl; repeat=!repeat; } result<<=1; result|=(c-'0'); } if(!repeat) { if(result<min result>max) { std::cout<<"The binary value must be in the decimal range "<<min<<".."<<max<<"\n"<<std::endl; repeat=!repeat; } } } return( result ); } int main() { uint binary = input_binary(0x1,0xffffff); uint temp=binary; uint len=1; while(temp>>=3) ++len; std::string octal(len, '0'); for( std::string::reverse_iterator i=octal.rbegin(); i!=octal.rend(); ++i ) { *i=(binary & 0x7)+'0'; binary>>=3; } std::cout<<"Octal: "<<octal.c_str()<<"\n"<<std::endl; return(0); }
In the US, inches. In metric countries, CMs.
10 to the twenty-third power is not a unit of measurement. It is a number.
not possible to convert from a length uint (Mile) to an area unit (sq. ft.)
cause of a fuel leak in the multi port uint ,broken tubing usually.
not possible to convert from a length uint (Mile) to an area unit (sq. ft.)