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《OFFER14》14_CuttingRope

时间:2018-12-08 13:05:18      阅读:117      评论:0      收藏:0      [点我收藏+]

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  1 // 面试题14:剪绳子
  2 // 题目:给你一根长度为n绳子,请把绳子剪成m段(m、n都是整数,n>1并且m≥1)。
  3 // 每段的绳子的长度记为k[0]、k[1]、……、k[m]。k[0]*k[1]*…*k[m]可能的最大乘
  4 // 积是多少?例如当绳子的长度是8时,我们把它剪成长度分别为2、3、3的三段,此
  5 // 时得到最大的乘积18。
  6 
  7 #include <iostream>
  8 #include <cmath>
  9 
 10 // ====================动态规划====================
 11 int maxProductAfterCutting_solution1(int length)
 12 {
 13     if(length < 2)
 14         return 0;
 15     if(length == 2)
 16         return 1;
 17     if(length == 3)
 18         return 2;
 19 
 20     int* products = new int[length + 1];
 21     products[0] = 0;
 22     products[1] = 1;
 23     products[2] = 2;
 24     products[3] = 3;
 25 
 26     int max = 0;
 27     for(int i = 4; i <= length; ++i)
 28     {
 29         max = 0;
 30         for(int j = 1; j <= i / 2; ++j)
 31         {
 32             int product = products[j] * products[i - j];
 33             if(max < product)
 34                 max = product;
 35 
 36             products[i] = max;
 37         }
 38     }
 39 
 40     max = products[length];
 41     delete[] products;
 42 
 43     return max;
 44 }
 45 
 46 // ====================贪婪算法====================
 47 int maxProductAfterCutting_solution2(int length)
 48 {
 49     if(length < 2)
 50         return 0;
 51     if(length == 2)
 52         return 1;
 53     if(length == 3)
 54         return 2;
 55 
 56     // 尽可能多地减去长度为3的绳子段
 57     int timesOf3 = length / 3;
 58 
 59     // 当绳子最后剩下的长度为4的时候,不能再剪去长度为3的绳子段。
 60     // 此时更好的方法是把绳子剪成长度为2的两段,因为2*2 > 3*1。
 61     if(length - timesOf3 * 3 == 1)
 62         timesOf3 -= 1;
 63 
 64     int timesOf2 = (length - timesOf3 * 3) / 2;
 65 
 66     return (int) (pow(3, timesOf3)) * (int) (pow(2, timesOf2));
 67 }
 68 
 69 // ====================测试代码====================
 70 void test(const char* testName, int length, int expected)
 71 {
 72     int result1 = maxProductAfterCutting_solution1(length);
 73     if(result1 == expected)
 74         std::cout << "Solution1 for " << testName << " passed." << std::endl;
 75     else
 76         std::cout << "Solution1 for " << testName << " FAILED." << std::endl;
 77 
 78     int result2 = maxProductAfterCutting_solution2(length);
 79     if(result2 == expected)
 80         std::cout << "Solution2 for " << testName << " passed." << std::endl;
 81     else
 82         std::cout << "Solution2 for " << testName << " FAILED." << std::endl;
 83 }
 84 
 85 void test1()
 86 {
 87     int length = 1;
 88     int expected = 0;
 89     test("test1", length, expected);
 90 }
 91 
 92 void test2()
 93 {
 94     int length = 2;
 95     int expected = 1;
 96     test("test2", length, expected);
 97 }
 98 
 99 void test3()
100 {
101     int length = 3;
102     int expected = 2;
103     test("test3", length, expected);
104 }
105 
106 void test4()
107 {
108     int length = 4;
109     int expected = 4;
110     test("test4", length, expected);
111 }
112 
113 void test5()
114 {
115     int length = 5;
116     int expected = 6;
117     test("test5", length, expected);
118 }
119 
120 void test6()
121 {
122     int length = 6;
123     int expected = 9;
124     test("test6", length, expected);
125 }
126 
127 void test7()
128 {
129     int length = 7;
130     int expected = 12;
131     test("test7", length, expected);
132 }
133 
134 void test8()
135 {
136     int length = 8;
137     int expected = 18;
138     test("test8", length, expected);
139 }
140 
141 void test9()
142 {
143     int length = 9;
144     int expected = 27;
145     test("test9", length, expected);
146 }
147 
148 void test10()
149 {
150     int length = 10;
151     int expected = 36;
152     test("test10", length, expected);
153 }
154 
155 void test11()
156 {
157     int length = 50;
158     int expected = 86093442;
159     test("test11", length, expected);
160 }
161 
162 int main(int agrc, char* argv[])
163 {
164     test1();
165     test2();
166     test3();
167     test4();
168     test5();
169     test6();
170     test7();
171     test8();
172     test9();
173     test10();
174     test11();
175 
176     return 0;
177 }
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《OFFER14》14_CuttingRope

标签:isp   res   rop   nbsp   sof   lse   算法   ons   ret   

原文地址:https://www.cnblogs.com/focus-z/p/10086929.html

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