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64_Accumulate.cpp
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/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题64：求1+2+…+n
// 题目：求1+2+…+n，要求不能使用乘除法、for、while、if、else、switch、case
// 等关键字及条件判断语句（A?B:C）。
#include <cstdio>
// ====================方法一====================
class Temp
{
public:
Temp() { ++ N; Sum += N; }
static void Reset() { N = 0; Sum = 0; }
static unsigned int GetSum() { return Sum; }
private:
static unsigned int N;
static unsigned int Sum;
};
unsigned int Temp::N = 0;
unsigned int Temp::Sum = 0;
unsigned int Sum_Solution1(unsigned int n)
{
Temp::Reset();
Temp *a = new Temp[n];
delete []a;
a = NULL;
return Temp::GetSum();
}
// ====================方法二====================
class A;
A* Array[2];
class A
{
public:
virtual unsigned int Sum (unsigned int n)
{
return 0;
}
};
class B: public A
{
public:
virtual unsigned int Sum (unsigned int n)
{
return Array[!!n]->Sum(n-1) + n; //把数值变量n转换为bool类型，对n连续做两次反运算，那么非0的n转换为true，0转换为false
}
};
int Sum_Solution2(int n)
{
A a;
B b;
Array[0] = &a;
Array[1] = &b;
int value = Array[1]->Sum(n);
return value;
}
// ====================方法三====================
typedef unsigned int (*fun)(unsigned int);
unsigned int Solution3_Teminator(unsigned int n)
{
return 0;
}
unsigned int Sum_Solution3(unsigned int n)
{
static fun f[2] = {Solution3_Teminator, Sum_Solution3};
return n + f[!!n](n - 1);
}
// ====================方法四====================
template <unsigned int n> struct Sum_Solution4
{
enum Value { N = Sum_Solution4<n - 1>::N + n};
};
template <> struct Sum_Solution4<1>
{
enum Value { N = 1};
};
template <> struct Sum_Solution4<0>
{
enum Value { N = 0};
};
// ====================方法五====================
unsigned int Sum_Solution5(unsigned int n)
{
unsigned int sum = n;
//bool b = (n == 0) || (sum += Sum_Solution5(n - 1)) > 0;
bool b = (n > 0) && (sum += Sum_Solution5(n - 1)) > 0;
return sum;
}
// ====================测试代码====================
void Test(int n, int expected)
{
printf("Test for %d begins:\n", n);
if(Sum_Solution1(n) == expected)
printf("Solution1 passed.\n");
else
printf("Solution1 failed.\n");
if(Sum_Solution2(n) == expected)
printf("Solution2 passed.\n");
else
printf("Solution2 failed.\n");
if(Sum_Solution3(n) == expected)
printf("Solution3 passed.\n");
else
printf("Solution3 failed.\n");
if(Sum_Solution5(n) == expected)
printf("Solution5 passed.\n");
else
printf("Solution5 failed.\n");
}
void Test1()
{
const unsigned int number = 1;
int expected = 1;
Test(number, expected);
if(Sum_Solution4<number>::N == expected)
printf("Solution4 passed.\n");
else
printf("Solution4 failed.\n");
}
void Test2()
{
const unsigned int number = 5;
int expected = 15;
Test(number, expected);
if(Sum_Solution4<number>::N == expected)
printf("Solution4 passed.\n");
else
printf("Solution4 failed.\n");
}
void Test3()
{
const unsigned int number = 10;
int expected = 55;
Test(number, expected);
if(Sum_Solution4<number>::N == expected)
printf("Solution4 passed.\n");
else
printf("Solution4 failed.\n");
}
void Test4()
{
const unsigned int number = 0;
int expected = 0;
Test(number, expected);
if(Sum_Solution4<number>::N == expected)
printf("Solution4 passed.\n");
else
printf("Solution4 failed.\n");
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
return 0;
}