see the program
The person or program that solves the equation does.
write a c++program by using if statement to read a number and check whether it is positive or negative
$n = 10*(1+10)/2;
By learning how to program on C+.
prompt x floor(x + .5) -> x disp x
Just generate the Fibonacci numbers one by one, and print each number's last digit ie number%10.
write a program that reads in the size of the side of square and then pints a hollow square of that size out of asterisks and blanks?
The Fibonacci sequence uses recursion to derive answers. It is defined as: F0 = 0 F1 = 1 Fn = F(n - 1) + F(n -2) To have this sequence printed by a php script use the following: function fibonacci($n) { if($n 1) return 1; //F1 else return fibonacci($n - 1) + fibonacci($n - 2); //Fn } This recursive function will print out the Fibonacci number for the integer n. To make it print out all the numbers in a particular set add this to your script. for($i = 0; $i < 15; $i++) { echo fibonacci($i) . "<br />"; } So your final result would look like. <?php function fibonacci($n) { if($n 1) return 1; else return fibonacci($n - 1) + fibonacci($n - 2); } for($i = 0; $i < 15; $i++) { echo fibonacci($i) . "<br />"; } ?>
Exactly what do you mean by 'C program in Java'
x -=y;
You mean you have written a program, but you don't understand it? Well, how could I explain it without seeing it?
1 hour ago my c program said no, but now I know 5 actually is an integer!
i dn't know. haha
Output a prompt.Either:Read from standard input (std::cin) to an integer.Or:Read a line from standard input (std::getline()) to a string.Create a string stream (std::stringstream) to read the string.Read from the string stream to an integer.For each integer from 2 to half the entered integer:If the entered integer is divisible by the current integer:The number is not prime.Exit the program.The number is prime.Exit the program.
it would be a simple loop... something like dim counter as integer dim f1 as integer dim f2 as integer dim f3 as integer counter = 0 f1 = 0 f2 = 1 listbox.additem(f1) listbox.additem(f2) While (count <= 610) ( f3 = f2+f1 f1=f2 f2=f3 listbox.additem(f3) ) next count Those lines of code aren't purfect (you can't copy and paste) but the idea behind the loop will be the same. This will out put the numbers into a list box. If you want to do it in like... a message box, you could do: dim answer as string answer = f1 &" " & f2 while ( ) answer = answer & " " & f3 next count msgbox.show("The first 610 Fibonacci numbers are: " & answer) Shoot me any questions via email @ kjhansen1@cox.net
#include<iostream> #include<sstream> using namespace std; unsigned sum_of_squares (const unsigned max) { if (max==0) return 0; if (max==1) return 1; return sum_of_squares (max-1) + (max*max); } int main () { unsigned num = 0; while (1) { cout << "Enter a positive integer (0 to exit): "; string s; cin >> s; if (s[0]=='0') break; stringstream ss; ss << s; if (ss >> num) { cout << "The sum of all squares from 1 to " << num << " is: " << sum_of_squares (num) << endl; continue; } cerr << "Invalid input: " << s << endl; } cout << "Quitting..." << endl; } Example output: Enter a positive integer (0 to exit): 1 The sum of all squares from 1 to 1 is: 1 Enter a positive integer (0 to exit): 2 The sum of all squares from 1 to 2 is: 5 Enter a positive integer (0 to exit): 3 The sum of all squares from 1 to 3 is: 14 Enter a positive integer (0 to exit): 4 The sum of all squares from 1 to 4 is: 30 Enter a positive integer (0 to exit): 5 The sum of all squares from 1 to 5 is: 55 Enter a positive integer (0 to exit): 6 The sum of all squares from 1 to 6 is: 91 Enter a positive integer (0 to exit): 7 The sum of all squares from 1 to 7 is: 140 Enter a positive integer (0 to exit): 0 Quitting...
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