剪绳子

问题：给你一根长度为n的绳子，请把绳子剪成m段（m,n都是整数，n>1,m>1），每段绳子的长度记为k[0],k[1],...,k[m]。请问k[0]*k[1]*...k[m]可能的最大乘积是多少？

#include<iostream>#include<math.h>using namespace std;int maxProductAfterCutting_solution(int length);int maxProductAfterCutting_solution2(int length);int main(){cout <<"0: "<< maxProductAfterCutting_solution(0) <<endl;cout <<"1: "<< maxProductAfterCutting_solution(1) << endl;cout <<"2: "<< maxProductAfterCutting_solution(2) << endl;cout <<"3: "<< maxProductAfterCutting_solution(3) << endl;cout <<"4: "<< maxProductAfterCutting_solution(4) << endl;cout <<"5: "<< maxProductAfterCutting_solution(5) << endl;cout <<"6: "<< maxProductAfterCutting_solution(6) << endl;cout <<"7: "<< maxProductAfterCutting_solution(7) << endl;cout << "8: " << maxProductAfterCutting_solution(8) << endl;cout << "9: " << maxProductAfterCutting_solution(9) << endl;cout << "10: " << maxProductAfterCutting_solution(10) << endl;cout << "11: " << maxProductAfterCutting_solution(11) << endl;system("pause");    return 0;}int maxProductAfterCutting_solution(int length) {if (length < 2) {return 0;}if (length==2){return 2;}if (length ==3){return 3;}int *products = new int[length +1];products[0] = 0;products[1] = 1;products[2] = 2;products[3] = 3;int max = 0;for (size_t i = 4; i <=length; i++){int mid = i / 2;max = 0;for (size_t j = 1; j <= mid; j++){int tmp = products[j] * products[i - j];if (max< tmp){max = tmp;}products[i] = max;}}max = products[length];delete[] products;return max;}int maxProductAfterCutting_solution2(int length) {if (length<2){return 0;}if (length ==2){return 2;}if (length == 3){return 3;}int time3 = length / 3;int time4 = 0;if (length%3 ==1){time3 = time3 - 1;time4 = 1;}int time2 = (length - 3 * time3)/2;return (pow(3, time3) * pow(2, time2));}