You will NEVER get a pop star's telephone number on here - They keep their personal numbers VERY private - to stop being pestered by fans. The only way you'll be able to contact her - is through her agent or record company. Think about it - if YOU were in her position - would YOU want to get hundreds, perhaps thousands of phone calls a day from fans ?
this is the script Harry: I,m harry potter, school is for losers, I,m totally awsome Snape: Harry Potter, you've been absent from potions class for 3 weeks and I have no choice but to (goes muffled)num num num num num num Harry Potter num num num (harry punches snape then plays saxaphone) awsome
Wikianswers does not give out celebrities personal phone numbers.
what is kiely Williams number
The link for verilog question paper is http://www.interview-secrets.net/jobinterviews/verilog-interview-questions.html http://vlsifaq.blogspot.com/2007/10/verilog.html http://vlsifaq.blogspot.com/2007/10/verilog.html http://forum.rficdesign.com/YaBB.pl?num=1222165121 http://forum.rficdesign.com/YaBB.pl?num=1222165286 http://www.asicguru.com/system-verilog/interview-questions/10/
Simply press the button called 'Num Lock'. Toggling it on and off locks and unlocks the use of the numbers on the right hand side of pretty much ANY computer keyboard, not just laptops. If that's not what you're after, some keyboards have a 'fn' or 'f' lock. Try that one too.
Such a method could be: public int NumberOfDigits(int num) { num = num/10; if( num == 0) return 1; return ( 1+NumberOfDigits(num) ); }
shilpa anand give me your cell num
unsigned binary_to_gray (unsigned num) { return num ^ (num >> 1); } unsigned gray_to_binary (unsigned num) { /* note: assumes num is no more than 32-bits in length */ num ^= (num >> 16); num ^= (num >> 8); num ^= (num >> 4); num ^= (num >> 2); num ^= (num >> 1); return num ; }
there could be a part in it like this: int num, digit; int count [10]; do { digit = num%10; num != 10; ++count[digit]; } while (num);
#include<iostream> #include<iomanip> #include<math.h> // for floor, log and pow bool is_fib(unsigned num) { // Algorithm: num is Fibonacci if 5*num*num+4 or 5*num*num-4 is a perfect square. // Note: // Since 5*num*num could result in a huge number which could easily overflow, // we'll use a modified algorithm that utilises numbers smaller than num. double root5 = std::sqrt(5.0); double phi = (1 + root5) / 2; long idx = (long)floor( log(num*root5) / log(phi) + 0.5 ); long n = (long)floor( pow(phi, idx)/root5 + 0.5); return (n == num); } int main() { // Print non-Fibonacci numbers in range: 0-50 for (unsigned num=0; num<=50; ++num) if (!is_fib(num)) std::cout <<std::setw(6) <<num <<" is not a Fibonacci number" <<std::endl; }
num num num
here 9273482784727490
I will give you jilly.Her BFF is num nums.
In simple Python code: def convertToAngle(num): while True: if num < 0: num = 360 - num elif num > 360: num -= 360 else: break return num
#include<iosys.h> #include<math.h> // for sqrt() bool is_prime (unsigned num) { if (num<2) return false; // 2 is the first prime if (!(num%2)) return num==2; // 2 is the only even prime // largest potential factor is square root of num unsigned max = (unsigned) sqrt ((double) num) + 1; // test all odd factors for (unsigned factor=3; factor<max; factor+=2) if (!(num%factor)) return false; return true; // if we get this far, num has no factors and is therefore prime } int main (void) { // test all nums from 0 to 100 inclusive for (unsigned num=0; num<=100; ++num) { if (is_prime (num)) printf ("%d is prime\n", num); else if (num>0) printf ("%d is composite\n", num); else printf ("%d is neither prime nor composite\n", %d); } return 0; }
#include<iostream> #include<sstream> #include<exception> std::string decimal_to_roman (unsigned num) { std::stringstream ss {}; while (num>0) { if (num>10000) throw std::range_error ( "ERROR: decimal_to_roman (unsigned num) [num is out of range]"); else if (num==10000) { ss<<"[M]"; num-=10000; } else if (num>=9000) { ss<<"[CM]"; num-=9000; } else if (num>=5000) { ss<<"[D]"; num-=5000; } else if (num>=4000) { ss<<"[CD]"; num-=4000; } else if (num>=1000) { ss<<"M"; num-=1000; } else if (num>=900) { ss<<"CM"; num-=900; } else if (num>=500) { ss<<"D"; num-=500; } else if (num>=400) { ss<<"CD"; num-=400; } else if (num>=100) { ss<<"C"; num-=100; } else if (num>=90) { ss<<"XC"; num-=90; } else if (num>=50) { ss<<"L"; num-=50; } else if (num>=40) { ss<<"XL"; num-=40; } else if (num>=10) { ss<<"X"; num-=10; } else if (num==9) { ss<<"IX"; num-=9; } else if (num>=5) { ss<<"V"; num-=5; } else if (num==4) { ss<<"IV"; num-=4; } else if (num>=1) { ss<<"I"; num-=1; } } return ss.str(); } int main (void) { for (unsigned n=1; n<=10000; ++n) { try { std::cout << n << "\t = " << decimal_to_roman(n) << std::endl; } catch (std::range_error& e) { std::cerr<<e.what()<<std::endl; break; } } }
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