JZ-C-46

Posted 2020-07-31 回看欧洲

剑指offer第四十六题：求1+2+3+...+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）

  1 //============================================================================
  2 // Name        : JZ-C-46.cpp
  3 // Author      : Laughing_Lz
  4 // Version     :
  5 // Copyright   : All Right Reserved
  6 // Description : 求1+2+3+...+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）
  7 //============================================================================
  8 
  9 #include <iostream>
 10 #include <stdio.h>
 11 using namespace std;
 12 // ====================方法一:利用构造函数求解====================
 13 class Temp {
 14 public:
 15     Temp() {
 16         ++N;
 17         Sum += N;
 18     }
 19 
 20     static void Reset() { //注意这里是静态方法
 21         N = 0;
 22         Sum = 0;
 23     }
 24     static unsigned int GetSum() { //注意这里是静态方法
 25         return Sum;
 26     }
 27 
 28 private:
 29     static unsigned int N; //注意这里是静态变量
 30     static unsigned int Sum; //注意这里是静态变量
 31 };
 32 
 33 unsigned int Temp::N = 0;
 34 unsigned int Temp::Sum = 0;
 35 
 36 unsigned int Sum_Solution1(unsigned int n) {
 37     Temp::Reset();
 38 
 39     Temp *a = new Temp[n]; //这里创建的是对象数组
 40     delete[] a;
 41     a = NULL;
 42 
 43     return Temp::GetSum();
 44 }
 45 
 46 // ====================方法二:利用虚函数求解====================
 47 /**
 48  * 一个函数充当递归函数的角色，另一个函数处理终止递归的情况，因此使用布尔变量，若n不为0，!!n为1;若n为0，!!n为0；
 49  */
 50 class A;
 51 A* Array[2];
 52 
 53 class A {
 54 public:
 55     virtual unsigned int Sum(unsigned int n) {
 56         return 0;
 57     }
 58 };
 59 
 60 class B: public A {
 61 public:
 62     virtual unsigned int Sum(unsigned int n) {
 63         return Array[!!n]->Sum(n - 1) + n;
 64     }
 65 };
 66 
 67 int Sum_Solution2(int n) {
 68     A a;
 69     B b;
 70     Array[0] = &a; //递归结束，调用的为父类Sum方法
 71     Array[1] = &b;
 72 
 73     int value = Array[1]->Sum(n);
 74 
 75     return value;
 76 }
 77 
 78 // ====================方法三：利用函数指针求解====================
 79 typedef unsigned int (*fun)(unsigned int); //函数指针定义
 80 
 81 unsigned int Solution3_Teminator(unsigned int n) {
 82     return 0;
 83 }
 84 
 85 unsigned int Sum_Solution3(unsigned int n) {
 86     static fun f[2] = { Solution3_Teminator, Sum_Solution3 };
 87     return n + f[!!n](n - 1);
 88 }
 89 
 90 // ====================方法四：利用模板类型求解====================
 91 template<unsigned int n> struct Sum_Solution4 {
 92     enum Value {//enum：枚举类型
 93         N = Sum_Solution4<n - 1>::N + n
 94     };
 95 };
 96 
 97 template<> struct Sum_Solution4<1> {
 98     enum Value {
 99         N = 1
100     };
101 };
102 
103 template<> struct Sum_Solution4<0> {
104     enum Value {
105         N = 0
106     };
107 };
108 
109 // ====================测试代码====================
110 void Test(int n, int expected) {
111     printf("Test for %d begins:\n", n);
112 
113     if (Sum_Solution1(n) == expected)
114         printf("Solution1 passed.\n");
115     else
116         printf("Solution1 failed.\n");
117 
118     if (Sum_Solution2(n) == expected)
119         printf("Solution2 passed.\n");
120     else
121         printf("Solution2 failed.\n");
122 
123     if (Sum_Solution3(n) == expected)
124         printf("Solution3 passed.\n");
125     else
126         printf("Solution3 failed.\n");
127 }
128 
129 void Test1() {
130     const unsigned int number = 1;
131     int expected = 1;
132     Test(number, expected);
133     if (Sum_Solution4<number>::N == expected)
134         printf("Solution4 passed.\n");
135     else
136         printf("Solution4 failed.\n");
137 }
138 
139 void Test2() {
140     const unsigned int number = 5;
141     int expected = 15;
142     Test(number, expected);
143     if (Sum_Solution4<number>::N == expected)
144         printf("Solution4 passed.\n");
145     else
146         printf("Solution4 failed.\n");
147 }
148 
149 void Test3() {
150     const unsigned int number = 10;
151     int expected = 55;
152     Test(number, expected);
153     if (Sum_Solution4<number>::N == expected)
154         printf("Solution4 passed.\n");
155     else
156         printf("Solution4 failed.\n");
157 }
158 
159 void Test4() {
160     const unsigned int number = 0;
161     int expected = 0;
162     Test(number, expected);
163     if (Sum_Solution4<number>::N == expected)
164         printf("Solution4 passed.\n");
165     else
166         printf("Solution4 failed.\n");
167 }
168 
169 int main(int argc, char** argvb) {
170     Test1();
171     Test2();
172     Test3();
173     Test4();
174 
175     return 0;
176 }