yo yo as
Chat with our AI personalities
#include<iostream> using namespace std; int main(void) { int a; int b; { cout << "Please enter a number: "; cin >> a; b = a%2 == 0 ? a*a : a*a*a; cout << "The result is: " << b << endl; } return 0; } - - - //Alternative #include<iostream> #include<cmath>
# include # include { float centi;farha; clrscr( ); print f ("Enter the temp); scan f ("%f"&centi); farha(a*centi)3/5; print f("farhanhite=%2f"farhan); getch( ); }
it's easy i am assuming you know programming as you are asking a question of a high standard. The formula for any number in the traingle is (r-1)!/[(c-1)!*(r-c)!] where r represents row number and c represents column number. Note:- '!' sign means factorial. Eg:- 5! = 5 x 4 x 3 x 2 x1.
400-500 years now for 2009 duby duby doo wap Phineas and ferb = Candice and perr y
First check if it is divisible by 13. You need to delete the last digit from the number, then subtract 9 times the deleted digit from the remaining number. If what is left is divisible by 13, then so is the original number. For example. 195 The last digit is 5 so we delete that. Now 9x5=45 and we must subtract that from 195 So we have 19-45=-26 which is clearly divisible by 13. Now for 11 you need another test. Alternately add and subtract the digits from left to right. (You can think of the first digit as being 'added' to zero.) If the result (including 0) is divisible by 11, the number is also. Example: to see whether 365167484 is divisible by 11, start by subtracting: [0+]3-6+5-1+6-7+4-8+4 = 0; therefore 365167484 is divisible by 11. If your numbers passes both the divisibility tests, for 11 and 13, then it is divisible by both.