算法导论 第十五章 15-10 投资策略规划 O(n^2)

 save[j][i] = max((biggest - cost_2)*input[j][i], (save[j- 1][i] - cost_1)*input[j][i]);

#include <iostream>
#include <algorithm>

using namespace std;

void planning(double **input, int sizeI, int sizeJ, double cost_1, double cost_2);

int main() {
	double **input = new double *[5];
	for (int count = 0; count < 5; ++count) {
		input[count] = new double[5];
		for (int countJ = 0; countJ < 5; ++countJ)
			cin >> input[count][countJ];
	}
	planning(input, 5, 5, 1, 2);
}

void planning(double **input, int sizeI, int sizeJ, double cost_1, double cost_2) { //sizeI年，每年sizeJ个选择
	double **save = new double *[sizeJ];
	for (int count = 0; count < sizeJ; ++count)
		save[count] = new double[sizeI];
	double biggest = 0;								//这里的biggest和index分别记当前年数的最大值和其索引
	int index = -1;									
	for (int count = 0; count < sizeI; ++count) {
		save[0][count] = input[0][count];
		if (save[0][count] > biggest) {
			biggest = save[0][count];