// Copyright 2005, Google Inc. // All rights reserved. // // Redistribution and use in source and binary forms, with or without // modification, are permitted provided that the following conditions are // met: // // * Redistributions of source code must retain the above copyright // notice, this list of conditions and the following disclaimer. // * Redistributions in binary form must reproduce the above // copyright notice, this list of conditions and the following disclaimer // in the documentation and/or other materials provided with the // distribution. // * Neither the name of Google Inc. nor the names of its // contributors may be used to endorse or promote products derived from // this software without specific prior written permission. // // THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS // "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT // LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR // A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT // OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, // SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT // LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, // DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY // THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT // (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE // OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. // A sample program demonstrating using Google C++ testing framework. // // Author: wan@google.com (Zhanyong Wan) #include "sample1.h" // Returns n! (the factorial of n). For negative n, n! is defined to be 1. int Factorial(int n) { int result = 1; for (int i = 1; i <= n; i++) { result *= i; } return result; } // Returns true iff n is a prime number. bool IsPrime(int n) { // Trivial case 1: small numbers if (n <= 1) return false; // Trivial case 2: even numbers if (n % 2 == 0) return n == 2; // Now, we have that n is odd and n >= 3. // Try to divide n by every odd number i, starting from 3 for (int i = 3; ; i += 2) { // We only have to try i up to the squre root of n if (i > n/i) break; // Now, we have i <= n/i < n. // If n is divisible by i, n is not prime. if (n % i == 0) return false; } // n has no integer factor in the range (1, n), and thus is prime. return true; }