《剑指offer》第十四题（剪绳子）

// 面试题：剪绳子
// 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）。
// 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
// 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
// 时得到最大的乘积18。

#include <iostream>
#include <cmath>

// ====================动态规划====================
//四个特点;
//求问题最优解
//问题最优解可以分解为子问题的最优解
//可以分解为具有重复的子问题
//从上向下分析问题，从下向上计算问题

int maxProductAfterCutting_solution1(int length)
{
    if (length < 2)//当大小小于3的时候可以直接得出
        return 0;
    if (length == 2)
        return 1;
    if (length == 3)
        return 2;

    int* products = new int[length + 1];
    products[0] = 0;
    products[1] = 1;
    products[2] = 2;
    products[3] = 3;

    int max = 0;
    for (int i = 4; i <= length; ++i)
    {
        max = 0;
        for (int j = 1; j <= i / 2; ++j)
        {
            int product = products[j] * products[i - j];//products[j]和products[i - j]都是已经得到的
            if (max < product)
                max = product;

            products[i] = max;//选择存起来每个子问题的最优解，提高效率
        }
    }

    max = products[length];
    delete[] products;

    return max;
}

// ====================贪婪算法====================
//需要数学底子，比如下面这个就得公式证明3(n-3)>=2(n-2)，n>=5。。

int maxProductAfterCutting_solution2(int length)
{
    if (length < 2)
        return 0;
    if (length == 2)
        return 1;
    if (length == 3)
        return 2;

    // 尽可能多地减去长度为3的绳子段
    int timesOf3 = length / 3;

    // 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。
    // 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。
    if (length - timesOf3 * 3 == 1)
        timesOf3 -= 1;

    int timesOf2 = (length - timesOf3 * 3) / 2;

    return (int)(pow(3, timesOf3)) * (int)(pow(2, timesOf2));
}

// ====================测试代码====================
void test(const char* testName, int length, int expected)
{
    int result1 = maxProductAfterCutting_solution1(length);
    if (result1 == expected)
        std::cout << "Solution1 for " << testName << " passed." << std::endl;
    else
        std::cout << "Solution1 for " << testName << " FAILED." << std::endl;

    int result2 = maxProductAfterCutting_solution2(length);
    if (result2 == expected)
        std::cout << "Solution2 for " << testName << " passed." << std::endl;
    else
        std::cout << "Solution2 for " << testName << " FAILED." << std::endl;
}

void test1()
{
    int length = 1;
    int expected = 0;
    test("test1", length, expected);
}

void test2()
{
    int length = 2;
    int expected = 1;
    test("test2", length, expected);
}

void test3()
{
    int length = 3;
    int expected = 2;
    test("test3", length, expected);
}

void test4()
{
    int length = 4;
    int expected = 4;
    test("test4", length, expected);
}

void test5()
{
    int length = 5;
    int expected = 6;
    test("test5", length, expected);
}

void test6()
{
    int length = 6;
    int expected = 9;
    test("test6", length, expected);
}

void test7()
{
    int length = 7;
    int expected = 12;
    test("test7", length, expected);
}

void test8()
{
    int length = 8;
    int expected = 18;
    test("test8", length, expected);
}

void test9()
{
    int length = 9;
    int expected = 27;
    test("test9", length, expected);
}

void test10()
{
    int length = 10;
    int expected = 36;
    test("test10", length, expected);
}

void test11()
{
    int length = 50;
    int expected = 86093442;
    test("test11", length, expected);
}

int main(int agrc, char* argv[])
{
    test1();
    test2();
    test3();
    test4();
    test5();
    test6();
    test7();
    test8();
    test9();
    test10();
    test11();
    system("pause");
    return 0;
}

原文地址：https://www.cnblogs.com/CJT-blog/p/10480490.html