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// 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;
}
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