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